Python 实用宝典

Question 1

In python 3.4, I am typing

[] = ""

and it works fine, no Exception is raised. Though of course [] is not equal to "" afterwards.

[] = ()

also works fine.

"" = []

raises an exception as expected though,

() = ""

raises an exception as expected though. So, what’s going on?

Question 2

You are not comparing for equality. You are assigning.

Python allows you to assign to multiple targets:

foo, bar = 1, 2

assigns the two values to foo and bar, respectively. All you need is a sequence or iterable on the right-hand side, and a list or tuple of names on the left.

When you do:

[] = ""

you assigned an empty sequence (empty strings are sequences still) to an empty list of names.

It is essentially the same thing as doing:

[foo, bar, baz] = "abc"

where you end up with foo = "a", bar = "b" and baz = "c", but with fewer characters.

You cannot, however, assign to a string, so "" on the left-hand side of an assignment never works and is always a syntax error.

See the Assignment statements documentation:

An assignment statement evaluates the expression list (remember that this can be a single expression or a comma-separated list, the latter yielding a tuple) and assigns the single resulting object to each of the target lists, from left to right.

and

Assignment of an object to a target list, optionally enclosed in parentheses or square brackets, is recursively defined as follows.

Emphasis mine.

That Python doesn’t throw a syntax error for the empty list is actually a bit of a bug! The officially documented grammar doesn’t allow for an empty target list, and for the empty () you do get an error. See bug 23275; it is considered a harmless bug:

The starting point is recognizing that this has been around for very long time and is harmless.

Also see Why is it valid to assign to an empty list but not to an empty tuple?

Question 3

It follows the Assignment statements section rules from the documentation,

assignment_stmt ::=  (target_list "=")+ (expression_list | yield_expression)
If the target list is a comma-separated list of targets: The object must be an iterable with the same number of items as there are targets in the target list, and the items are assigned, from left to right, to the corresponding targets.

The object must be a sequence with the same number of items as there are targets in the target list, and the items are assigned, from left to right, to the corresponding targets.

So, when you say

[] = ""

"" is an iterable (any valid python string is an iterable) and it is being unpacked over the elements of the list.

For example,

>>> [a, b, c] = "123"
>>> a, b, c
('1', '2', '3')

Since you have an empty string, and an empty list, there is nothing to unpack. So, no error.

But, try this

>>> [] = "1"
Traceback (most recent call last):
  File "<input>", line 1, in <module>
ValueError: too many values to unpack (expected 0)
>>> [a] = ""
Traceback (most recent call last):
  File "<input>", line 1, in <module>
ValueError: need more than 0 values to unpack

In the [] = "1" case, you are trying to unpack the string "1" over an empty list of variables. So it complains with “too many values to unpack (expected 0)”.

Same way, in [a] = "" case, you have an empty string, so nothing to unpack really, but you are unpacking it over one variable, which is, again, not possible. That is why it complains “need more than 0 values to unpack”.

Apart from that, as you noticed,

>>> [] = ()

also throws no error, because () is an empty tuple.

>>> ()
()
>>> type(())
<class 'tuple'>

and when it is unpacked over an empty list, there is nothing to unpack. So no error.

But, when you do

>>> "" = []
  File "<input>", line 1
SyntaxError: can't assign to literal
>>> "" = ()
  File "<input>", line 1
SyntaxError: can't assign to literal

as the error message says, you are trying to assign to a string literal. Which is not possible. That is why you are getting the errors. It is like saying

>>> 1 = "one"
  File "<input>", line 1
SyntaxError: can't assign to literal

Internals

Internally, this assignment operation will be translated to UNPACK_SEQUENCE op code,

>>> dis(compile('[] = ""', "string", "exec"))
  1           0 LOAD_CONST               0 ('')
              3 UNPACK_SEQUENCE          0
              6 LOAD_CONST               1 (None)

Here, since the string is empty, UNPACK_SEQUENCE unpacks 0 times. But when you have something like this

>>> dis(compile('[a, b, c] = "123"', "string", "exec"))
  1           0 LOAD_CONST               0 ('123')
              3 UNPACK_SEQUENCE          3
              6 STORE_NAME               0 (a)
              9 STORE_NAME               1 (b)
             12 STORE_NAME               2 (c)
             15 LOAD_CONST               1 (None)
             18 RETURN_VALUE

the sequence 123 is unpacked in to the stack, from right to left. So, the top of the stack would be 1 and the next would be 2 and the last would be 3. Then it assigns from the top of the stack to the variables from the left hand side expression one by one.

BTW, in Python, this is how you can do multiple assignments in the same expression. For example,

a, b, c, d, e, f = u, v, w, x, y, z

this works because, the right hand values are used to construct a tuple and then it will be unpacked over the left hand side values.

>>> dis(compile('a, b, c, d, e, f = u, v, w, x, y, z', "string", "exec"))
  1           0 LOAD_NAME                0 (u)
              3 LOAD_NAME                1 (v)
              6 LOAD_NAME                2 (w)
              9 LOAD_NAME                3 (x)
             12 LOAD_NAME                4 (y)
             15 LOAD_NAME                5 (z)
             18 BUILD_TUPLE              6
             21 UNPACK_SEQUENCE          6
             24 STORE_NAME               6 (a)
             27 STORE_NAME               7 (b)
             30 STORE_NAME               8 (c)
             33 STORE_NAME               9 (d)
             36 STORE_NAME              10 (e)
             39 STORE_NAME              11 (f)
             42 LOAD_CONST               0 (None)
             45 RETURN_VALUE

but the classic swapping technique a, b = b, a uses rotation of elements in the top of the stack. If you have only two or three elements then they are treated with special ROT_TWO and ROT_THREE instructions instead of constructing the tuple and unpacking.

>>> dis(compile('a, b = b, a', "string", "exec"))
  1           0 LOAD_NAME                0 (b)
              3 LOAD_NAME                1 (a)
              6 ROT_TWO
              7 STORE_NAME               1 (a)
             10 STORE_NAME               0 (b)
             13 LOAD_CONST               0 (None)
             16 RETURN_VALUE

Question 4

I have a list of objects and I want to remove all objects that are empty except for one, using filter and a lambda expression.

For example if the input is:

[Object(name=""), Object(name="fake_name"), Object(name="")]

…then the output should be:

[Object(name=""), Object(name="fake_name")]

Is there a way to add an assignment to a lambda expression? For example:

flag = True 
input = [Object(name=""), Object(name="fake_name"), Object(name="")] 
output = filter(
    (lambda o: [flag or bool(o.name), flag = flag and bool(o.name)][0]),
    input
)

Question 5

The assignment expression operator := added in Python 3.8 supports assignment inside of lambda expressions. This operator can only appear within a parenthesized (...), bracketed [...], or braced {...} expression for syntactic reasons. For example, we will be able to write the following:

import sys
say_hello = lambda: (
    message := "Hello world",
    sys.stdout.write(message + "\n")
)[-1]
say_hello()

In Python 2, it was possible to perform local assignments as a side effect of list comprehensions.

import sys
say_hello = lambda: (
    [None for message in ["Hello world"]],
    sys.stdout.write(message + "\n")
)[-1]
say_hello()

However, it’s not possible to use either of these in your example because your variable flag is in an outer scope, not the lambda‘s scope. This doesn’t have to do with lambda, it’s the general behaviour in Python 2. Python 3 lets you get around this with the nonlocal keyword inside of defs, but nonlocal can’t be used inside lambdas.

There’s a workaround (see below), but while we’re on the topic…

In some cases you can use this to do everything inside of a lambda:

(lambda: [
    ['def'
        for sys in [__import__('sys')]
        for math in [__import__('math')]

        for sub in [lambda *vals: None]
        for fun in [lambda *vals: vals[-1]]

        for echo in [lambda *vals: sub(
            sys.stdout.write(u" ".join(map(unicode, vals)) + u"\n"))]

        for Cylinder in [type('Cylinder', (object,), dict(
            __init__ = lambda self, radius, height: sub(
                setattr(self, 'radius', radius),
                setattr(self, 'height', height)),

            volume = property(lambda self: fun(
                ['def' for top_area in [math.pi * self.radius ** 2]],

                self.height * top_area))))]

        for main in [lambda: sub(
            ['loop' for factor in [1, 2, 3] if sub(
                ['def'
                    for my_radius, my_height in [[10 * factor, 20 * factor]]
                    for my_cylinder in [Cylinder(my_radius, my_height)]],

                echo(u"A cylinder with a radius of %.1fcm and a height "
                     u"of %.1fcm has a volume of %.1fcm³."
                     % (my_radius, my_height, my_cylinder.volume)))])]],

    main()])()

A cylinder with a radius of 10.0cm and a height of 20.0cm has a volume of 6283.2cm³.
A cylinder with a radius of 20.0cm and a height of 40.0cm has a volume of 50265.5cm³.
A cylinder with a radius of 30.0cm and a height of 60.0cm has a volume of 169646.0cm³.

Please don’t.

…back to your original example: though you can’t perform assignments to the flag variable in the outer scope, you can use functions to modify the previously-assigned value.

For example, flag could be an object whose .value we set using setattr:

flag = Object(value=True)
input = [Object(name=''), Object(name='fake_name'), Object(name='')] 
output = filter(lambda o: [
    flag.value or bool(o.name),
    setattr(flag, 'value', flag.value and bool(o.name))
][0], input)

[Object(name=''), Object(name='fake_name')]

If we wanted to fit the above theme, we could use a list comprehension instead of setattr:

    [None for flag.value in [bool(o.name)]]

But really, in serious code you should always use a regular function definition instead of a lambda if you’re going to be doing outer assignment.

flag = Object(value=True)
def not_empty_except_first(o):
    result = flag.value or bool(o.name)
    flag.value = flag.value and bool(o.name)
    return result
input = [Object(name=""), Object(name="fake_name"), Object(name="")] 
output = filter(not_empty_except_first, input)

Question 6

You cannot really maintain state in a filter/lambda expression (unless abusing the global namespace). You can however achieve something similar using the accumulated result being passed around in a reduce() expression:

>>> f = lambda a, b: (a.append(b) or a) if (b not in a) else a
>>> input = ["foo", u"", "bar", "", "", "x"]
>>> reduce(f, input, [])
['foo', u'', 'bar', 'x']
>>>

You can, of course, tweak the condition a bit. In this case it filters out duplicates, but you can also use a.count(""), for example, to only restrict empty strings.

Needless to say, you can do this but you really shouldn’t. :)

Lastly, you can do anything in pure Python lambda: http://vanderwijk.info/blog/pure-lambda-calculus-python/

Question 7

There’s no need to use a lambda, when you can remove all the null ones, and put one back if the input size changes:

input = [Object(name=""), Object(name="fake_name"), Object(name="")] 
output = [x for x in input if x.name]
if(len(input) != len(output)):
    output.append(Object(name=""))

Question 8

Normal assignment (=) is not possible inside a lambda expression, although it is possible to perform various tricks with setattr and friends.

Solving your problem, however, is actually quite simple:

input = [Object(name=""), Object(name="fake_name"), Object(name="")]
output = filter(
    lambda o, _seen=set():
        not (not o and o in _seen or _seen.add(o)),
    input
    )

which will give you

[Object(Object(name=''), name='fake_name')]

As you can see, it’s keeping the first blank instance instead of the last. If you need the last instead, reverse the list going in to filter, and reverse the list coming out of filter:

output = filter(
    lambda o, _seen=set():
        not (not o and o in _seen or _seen.add(o)),
    input[::-1]
    )[::-1]

which will give you

[Object(name='fake_name'), Object(name='')]

One thing to be aware of: in order for this to work with arbitrary objects, those objects must properly implement __eq__ and __hash__ as explained here.

Question 9

UPDATE:

[o for d in [{}] for o in lst if o.name != "" or d.setdefault("", o) == o]

or using filter and lambda:

flag = {}
filter(lambda o: bool(o.name) or flag.setdefault("", o) == o, lst)

Previous Answer

OK, are you stuck on using filter and lambda?

It seems like this would be better served with a dictionary comprehension,

{o.name : o for o in input}.values()

I think the reason that Python doesn’t allow assignment in a lambda is similar to why it doesn’t allow assignment in a comprehension and that’s got something to do with the fact that these things are evaluated on the C side and thus can give us an increase in speed. At least that’s my impression after reading one of Guido’s essays.

My guess is this would also go against the philosophy of having one right way of doing any one thing in Python.

Question 10

TL;DR: When using functional idioms it’s better to write functional code

As many people have pointed out, in Python lambdas assignment is not allowed. In general when using functional idioms your better off thinking in a functional manner which means wherever possible no side effects and no assignments.

Here is functional solution which uses a lambda. I’ve assigned the lambda to fn for clarity (and because it got a little long-ish).

from operator import add
from itertools import ifilter, ifilterfalse
fn = lambda l, pred: add(list(ifilter(pred, iter(l))), [ifilterfalse(pred, iter(l)).next()])
objs = [Object(name=""), Object(name="fake_name"), Object(name="")]
fn(objs, lambda o: o.name != '')

You can also make this deal with iterators rather than lists by changing things around a little. You also have some different imports.

from itertools import chain, islice, ifilter, ifilterfalse
fn = lambda l, pred: chain(ifilter(pred, iter(l)), islice(ifilterfalse(pred, iter(l)), 1))

You can always reoganize the code to reduce the length of the statements.

Question 11

If instead of flag = True we can do an import instead, then I think this meets the criteria:

>>> from itertools import count
>>> a = ['hello', '', 'world', '', '', '', 'bob']
>>> filter(lambda L, j=count(): L or not next(j), a)
['hello', '', 'world', 'bob']

Or maybe the filter is better written as:

>>> filter(lambda L, blank_count=count(1): L or next(blank_count) == 1, a)

Or, just for a simple boolean, without any imports:

filter(lambda L, use_blank=iter([True]): L or next(use_blank, False), a)

Question 12

The pythonic way to track state during iteration is with generators. The itertools way is quite hard to understand IMHO and trying to hack lambdas to do this is plain silly. I’d try:

def keep_last_empty(input):
    last = None
    for item in iter(input):
        if item.name: yield item
        else: last = item
    if last is not None: yield last

output = list(keep_last_empty(input))

Overall, readability trumps compactness every time.

Question 13

No, you cannot put an assignment inside a lambda because of its own definition. If you work using functional programming, then you must assume that your values are not mutable.

One solution would be the following code:

output = lambda l, name: [] if l==[] \
             else [ l[ 0 ] ] + output( l[1:], name ) if l[ 0 ].name == name \
             else output( l[1:], name ) if l[ 0 ].name == "" \
             else [ l[ 0 ] ] + output( l[1:], name )

Question 14

If you need a lambda to remember state between calls, I would recommend either a function declared in the local namespace or a class with an overloaded __call__. Now that all my cautions against what you are trying to do is out of the way, we can get to an actual answer to your query.

If you really need to have your lambda to have some memory between calls, you can define it like:

f = lambda o, ns = {"flag":True}: [ns["flag"] or o.name, ns.__setitem__("flag", ns["flag"] and o.name)][0]

Then you just need to pass f to filter(). If you really need to, you can get back the value of flag with the following:

f.__defaults__[0]["flag"]

Alternatively, you can modify the global namespace by modifying the result of globals(). Unfortunately, you cannot modify the local namespace in the same way as modifying the result of locals() doesn’t affect the local namespace.

Question 15

You can use a bind function to use a pseudo multi-statement lambda. Then you can use a wrapper class for a Flag to enable assignment.

bind = lambda x, f=(lambda y: y): f(x)

class Flag(object):
    def __init__(self, value):
        self.value = value

    def set(self, value):
        self.value = value
        return value

input = [Object(name=""), Object(name="fake_name"), Object(name="")]
flag = Flag(True)
output = filter(
            lambda o: (
                bind(flag.value, lambda orig_flag_value:
                bind(flag.set(flag.value and bool(o.name)), lambda _:
                bind(orig_flag_value or bool(o.name))))),
            input)

Question 16

Kind of a messy workaround, but assignment in lambdas is illegal anyway, so it doesn’t really matter. You can use the builtin exec() function to run assignment from inside the lambda, such as this example:

>>> val
Traceback (most recent call last):
  File "<pyshell#31>", line 1, in <module>
    val
NameError: name 'val' is not defined
>>> d = lambda: exec('val=True', globals())
>>> d()
>>> val
True

Question 17

first , you dont need to use a local assigment for your job, just check the above answer

second, its simple to use locals() and globals() to got the variables table and then change the value

check this sample code:

print [locals().__setitem__('x', 'Hillo :]'), x][-1]

if you need to change the add a global variable to your environ, try to replace locals() with globals()

python’s list comp is cool but most of the triditional project dont accept this(like flask :[)

hope it could help

Question 18

import copy

a = "deepak"
b = 1, 2, 3, 4
c = [1, 2, 3, 4]
d = {1: 10, 2: 20, 3: 30}

a1 = copy.copy(a)
b1 = copy.copy(b)
c1 = copy.copy(c)
d1 = copy.copy(d)


print("immutable - id(a)==id(a1)", id(a) == id(a1))
print("immutable - id(b)==id(b1)", id(b) == id(b1))
print("mutable - id(c)==id(c1)", id(c) == id(c1))
print("mutable - id(d)==id(d1)", id(d) == id(d1))

I get the following results:

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) False
mutable - id(d)==id(d1) False

If I perform deepcopy:

a1 = copy.deepcopy(a)
b1 = copy.deepcopy(b)
c1 = copy.deepcopy(c)
d1 = copy.deepcopy(d)

results are the same:

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) False
mutable - id(d)==id(d1) False

If I work on assignment operations:

a1 = a
b1 = b
c1 = c
d1 = d

then results are:

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) True
mutable - id(d)==id(d1) True

Can somebody explain what exactly makes a difference between the copies? Is it something related to mutable & immutable objects? If so, can you please explain it to me?

Question 19

Normal assignment operations will simply point the new variable towards the existing object. The docs explain the difference between shallow and deep copies:

The difference between shallow and deep copying is only relevant for compound objects (objects that contain other objects, like lists or class instances):

A shallow copy constructs a new compound object and then (to the extent possible) inserts references into it to the objects found in the original.

A deep copy constructs a new compound object and then, recursively, inserts copies into it of the objects found in the original.

Here’s a little demonstration:

import copy

a = [1, 2, 3]
b = [4, 5, 6]
c = [a, b]

Using normal assignment operatings to copy:

d = c

print id(c) == id(d)          # True - d is the same object as c
print id(c[0]) == id(d[0])    # True - d[0] is the same object as c[0]

Using a shallow copy:

d = copy.copy(c)

print id(c) == id(d)          # False - d is now a new object
print id(c[0]) == id(d[0])    # True - d[0] is the same object as c[0]

Using a deep copy:

d = copy.deepcopy(c)

print id(c) == id(d)          # False - d is now a new object
print id(c[0]) == id(d[0])    # False - d[0] is now a new object

Question 20

For immutable objects, there is no need for copying because the data will never change, so Python uses the same data; ids are always the same. For mutable objects, since they can potentially change, [shallow] copy creates a new object.

Deep copy is related to nested structures. If you have list of lists, then deepcopy copies the nested lists also, so it is a recursive copy. With just copy, you have a new outer list, but inner lists are references.

Assignment does not copy. It simply sets the reference to the old data. So you need copy to create a new list with the same contents.

Question 21

For immutable objects, creating a copy don’t make much sense since they are not going to change. For mutable objects assignment,copy and deepcopy behaves differently. Lets talk about each of them with examples.

An assignment operation simply assigns the reference of source to destination e.g:

>>> i = [1,2,3]
>>> j=i
>>> hex(id(i)), hex(id(j))
>>> ('0x10296f908', '0x10296f908') #Both addresses are identical

Now i and j technically refers to same list. Both i and j have same memory address. Any updation to either of them will be reflected to the other. e.g:

>>> i.append(4)
>>> j
>>> [1,2,3,4] #Destination is updated

>>> j.append(5)
>>> i
>>> [1,2,3,4,5] #Source is updated

On the other hand copy and deepcopy creates a new copy of variable. So now changes to original variable will not be reflected to the copy variable and vice versa. However copy(shallow copy), don’t creates a copy of nested objects, instead it just copies the reference of nested objects. Deepcopy copies all the nested objects recursively.

Some examples to demonstrate behaviour of copy and deepcopy:

Flat list example using copy:

>>> import copy
>>> i = [1,2,3]
>>> j = copy.copy(i)
>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are different

>>> i.append(4)
>>> j
>>> [1,2,3] #Updation of original list didn't affected copied variable

Nested list example using copy:

>>> import copy
>>> i = [1,2,3,[4,5]]
>>> j = copy.copy(i)

>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are still different

>>> hex(id(i[3])), hex(id(j[3]))
>>> ('0x10296f908', '0x10296f908') #Nested lists have same address

>>> i[3].append(6)
>>> j
>>> [1,2,3,[4,5,6]] #Updation of original nested list updated the copy as well

Flat list example using deepcopy:

>>> import copy
>>> i = [1,2,3]
>>> j = copy.deepcopy(i)
>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are different

>>> i.append(4)
>>> j
>>> [1,2,3] #Updation of original list didn't affected copied variable

Nested list example using deepcopy:

>>> import copy
>>> i = [1,2,3,[4,5]]
>>> j = copy.deepcopy(i)

>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are still different

>>> hex(id(i[3])), hex(id(j[3]))
>>> ('0x10296f908', '0x102b9b7c8') #Nested lists have different addresses

>>> i[3].append(6)
>>> j
>>> [1,2,3,[4,5]] #Updation of original nested list didn't affected the copied variable

Question 22

Let’s see in a graphical example how the following code is executed:

import copy

class Foo(object):
    def __init__(self):
        pass


a = [Foo(), Foo()]
shallow = copy.copy(a)
deep = copy.deepcopy(a)

Question 23

a, b, c, d, a1, b1, c1 and d1 are references to objects in memory, which are uniquely identified by their ids.

An assignment operation takes a reference to the object in memory and assigns that reference to a new name. c=[1,2,3,4] is an assignment that creates a new list object containing those four integers, and assigns the reference to that object to c. c1=c is an assignment that takes the same reference to the same object and assigns that to c1. Since the list is mutable, anything that happens to that list will be visible regardless of whether you access it through c or c1, because they both reference the same object.

c1=copy.copy(c) is a “shallow copy” that creates a new list and assigns the reference to the new list to c1. c still points to the original list. So, if you modify the list at c1, the list that c refers to will not change.

The concept of copying is irrelevant to immutable objects like integers and strings. Since you can’t modify those objects, there is never a need to have two copies of the same value in memory at different locations. So integers and strings, and some other objects to which the concept of copying does not apply, are simply reassigned. This is why your examples with a and b result in identical ids.

c1=copy.deepcopy(c) is a “deep copy”, but it functions the same as a shallow copy in this example. Deep copies differ from shallow copies in that shallow copies will make a new copy of the object itself, but any references inside that object will not themselves be copied. In your example, your list has only integers inside it (which are immutable), and as previously discussed there is no need to copy those. So the “deep” part of the deep copy does not apply. However, consider this more complex list:

e = [[1, 2],[4, 5, 6],[7, 8, 9]]

This is a list that contains other lists (you could also describe it as a two-dimensional array).

If you run a “shallow copy” on e, copying it to e1, you will find that the id of the list changes, but each copy of the list contains references to the same three lists — the lists with integers inside. That means that if you were to do e[0].append(3), then e would be [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. But e1 would also be [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. On the other hand, if you subsequently did e.append([10, 11, 12]), e would be [[1, 2, 3],[4, 5, 6],[7, 8, 9],[10, 11, 12]]. But e1 would still be [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. That’s because the outer lists are separate objects that initially each contain three references to three inner lists. If you modify the inner lists, you can see those changes no matter if you are viewing them through one copy or the other. But if you modify one of the outer lists as above, then e contains three references to the original three lists plus one more reference to a new list. And e1 still only contains the original three references.

A ‘deep copy’ would not only duplicate the outer list, but it would also go inside the lists and duplicate the inner lists, so that the two resulting objects do not contain any of the same references (as far as mutable objects are concerned). If the inner lists had further lists (or other objects such as dictionaries) inside of them, they too would be duplicated. That’s the ‘deep’ part of the ‘deep copy’.

Question 24

In python, when we assign objects like list, tuples, dict, etc to another object usually with a ‘ = ‘ sign, python creates copy’s by reference. That is, let’s say we have a list of list like this :

list1 = [ [ 'a' , 'b' , 'c' ] , [ 'd' , 'e' , 'f' ]  ]

and we assign another list to this list like :

list2 = list1

then if we print list2 in python terminal we’ll get this :

list2 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f ']  ]

Both list1 & list2 are pointing to same memory location, any change to any one them will result in changes visible in both objects, i.e both objects are pointing to same memory location. If we change list1 like this :

list1[0][0] = 'x’
list1.append( [ 'g'] )

then both list1 and list2 will be :

list1 = [ [ 'x', 'b', 'c'] , [ 'd', 'e', ' f '] , [ 'g'] ]
list2 = [ [ 'x', 'b', 'c'] , [ 'd', 'e', ' f '] , [ 'g’ ] ]

Now coming to Shallow copy, when two objects are copied via shallow copy, the child object of both parent object refers to same memory location but any further new changes in any of the copied object will be independent to each other. Let’s understand this with a small example. Suppose we have this small code snippet :

import copy

list1 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f ']  ]      # assigning a list
list2 = copy.copy(list1)       # shallow copy is done using copy function of copy module

list1.append ( [ 'g', 'h', 'i'] )   # appending another list to list1

print list1
list1 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f '] , [ 'g', 'h', 'i'] ]
list2 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f '] ]

notice, list2 remains unaffected, but if we make changes to child objects like :

list1[0][0] = 'x’

then both list1 and list2 will get change :

list1 = [ [ 'x', 'b', 'c'] , [ 'd', 'e', ' f '] , [ 'g', 'h', 'i'] ] 
list2 = [ [ 'x', 'b', 'c'] , [ 'd', 'e', ' f '] ]

Now, Deep copy helps in creating completely isolated objects out of each other. If two objects are copied via Deep Copy then both parent & it’s child will be pointing to different memory location. Example :

import copy

list1 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f ']  ]         # assigning a list
list2 = deepcopy.copy(list1)       # deep copy is done using deepcopy function of copy module

list1.append ( [ 'g', 'h', 'i'] )   # appending another list to list1

print list1
list1 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f '] , [ 'g', 'h', 'i'] ]
list2 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f '] ]

notice, list2 remains unaffected, but if we make changes to child objects like :

list1[0][0] = 'x’

then also list2 will be unaffected as all the child objects and parent object points to different memory location :

list1 = [ [ 'x', 'b', 'c'] , [ 'd', 'e', ' f '] , [ 'g', 'h', 'i'] ] 
list2 = [ [ 'a', 'b', 'c'] , [ 'd', 'e', ' f  ' ] ]

Hope it helps.

Question 25

Below code demonstrates the difference between assignment, shallow copy using the copy method, shallow copy using the (slice) [:] and the deepcopy. Below example uses nested lists there by making the differences more evident.

from copy import deepcopy

########"List assignment (does not create a copy) ############
l1 = [1,2,3, [4,5,6], [7,8,9]]
l1_assigned = l1

print(l1)
print(l1_assigned)

print(id(l1), id(l1_assigned))
print(id(l1[3]), id(l1_assigned[3]))
print(id(l1[3][0]), id(l1_assigned[3][0]))

l1[3][0] = 100
l1.pop(4)
l1.remove(1)


print(l1)
print(l1_assigned)
print("###################################")

########"List copy using copy method (shallow copy)############

l2 = [1,2,3, [4,5,6], [7,8,9]]
l2_copy = l2.copy()

print(l2)
print(l2_copy)

print(id(l2), id(l2_copy))
print(id(l2[3]), id(l2_copy[3]))
print(id(l2[3][0]), id(l2_copy[3][0]))
l2[3][0] = 100
l2.pop(4)
l2.remove(1)


print(l2)
print(l2_copy)

print("###################################")

########"List copy using slice (shallow copy)############

l3 = [1,2,3, [4,5,6], [7,8,9]]
l3_slice = l3[:]

print(l3)
print(l3_slice)

print(id(l3), id(l3_slice))
print(id(l3[3]), id(l3_slice[3]))
print(id(l3[3][0]), id(l3_slice[3][0]))

l3[3][0] = 100
l3.pop(4)
l3.remove(1)


print(l3)
print(l3_slice)

print("###################################")

########"List copy using deepcopy ############

l4 = [1,2,3, [4,5,6], [7,8,9]]
l4_deep = deepcopy(l4)

print(l4)
print(l4_deep)

print(id(l4), id(l4_deep))
print(id(l4[3]), id(l4_deep[3]))
print(id(l4[3][0]), id(l4_deep[3][0]))

l4[3][0] = 100
l4.pop(4)
l4.remove(1)

print(l4)
print(l4_deep)
print("##########################")
print(l4[2], id(l4[2]))
print(l4_deep[3], id(l4_deep[3]))

print(l4[2][0], id(l4[2][0]))
print(l4_deep[3][0], id(l4_deep[3][0]))

Question 26

The GIST to take is this: Dealing with shallow lists (no sub_lists, just single elements) using “normal assignment” rises a “side effect” when you create a shallow list and then you create a copy of this list using “normal assignment”. This “side effect” is when you change any element of the copy list created, because it will automatically change the same elements of the original list. That is when copy comes in handy, as it won’t change the original list elements when changing the copy elements.

On the other hand, copy does have a “side effect” as well, when you have a list that has lists in it (sub_lists), and deepcopy solves it. For instance if you create a big list that has nested lists in it (sub_lists), and you create a copy of this big list (the original list). The “side effect” would arise when you modify the sub_lists of the copy list which would automatically modify the sub_lists of the big list. Sometimes (in some projects) you want to keep the big list (your original list) as it is without modification, and all you want is to make a copy of its elements (sub_lists). For that, your solution is to use deepcopy which will take care of this “side effect” and makes a copy without modifying the original content.

The different behaviors of copy and deep copy operations concerns only compound objects (ie: objects that contain other objects such as lists).

Here are the differences illustrated in this simple code example:

First

let’s check how copy (shallow) behaves, by creating an original list and a copy of this list:

import copy
original_list = [1, 2, 3, 4, 5, ['a', 'b']]
copy_list = copy.copy(original_list)

Now, let’s run some print tests and see how the original list behave compared to its copy list:

original_list and copy_list have different addresses

print(hex(id(original_list)), hex(id(copy_list))) # 0x1fb3030 0x1fb3328

elements of original_list and copy_list have the same addresses

print(hex(id(original_list[1])), hex(id(copy_list[1]))) # 0x537ed440 0x537ed440

sub_elements of original_list and copy_list have the same addresses

print(hex(id(original_list[5])), hex(id(copy_list[5]))) # 0x1faef08 0x1faef08

modifying original_list elements does NOT modify copy_list elements

original_list.append(6)
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b']]

modifying copy_list elements does NOT modify original_list elements

copy_list.append(7)
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b'], 7]

modifying original_list sub_elements automatically modify copy_list sub_elements

original_list[5].append('c')
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b', 'c'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b', 'c'], 7]

modifying copy_list sub_elements automatically modify original_list sub_elements

copy_list[5].append('d')
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b', 'c', 'd'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b', 'c', 'd'], 7]

Second

let’s check how deepcopy behaves, by doing the same thing as we did with copy (creating an original list and a copy of this list):

import copy
original_list = [1, 2, 3, 4, 5, ['a', 'b']]
copy_list = copy.copy(original_list)

Now, let’s run some print tests and see how the original list behave compared to its copy list:

import copy
original_list = [1, 2, 3, 4, 5, ['a', 'b']]
copy_list = copy.deepcopy(original_list)

original_list and copy_list have different addresses

print(hex(id(original_list)), hex(id(copy_list))) # 0x1fb3030 0x1fb3328

elements of original_list and copy_list have the same addresses

print(hex(id(original_list[1])), hex(id(copy_list[1]))) # 0x537ed440 0x537ed440

sub_elements of original_list and copy_list have different addresses

print(hex(id(original_list[5])), hex(id(copy_list[5]))) # 0x24eef08 0x24f3300

modifying original_list elements does NOT modify copy_list elements

original_list.append(6)
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b']]

modifying copy_list elements does NOT modify original_list elements

copy_list.append(7)
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b'], 7]

modifying original_list sub_elements does NOT modify copy_list sub_elements

original_list[5].append('c')
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b', 'c'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b'], 7]

modifying copy_list sub_elements does NOT modify original_list sub_elements

copy_list[5].append('d')
print("original_list is:", original_list) # original_list is: [1, 2, 3, 4, 5, ['a', 'b', 'c', 'd'], 6]
print("copy_list is:", copy_list) # copy_list is: [1, 2, 3, 4, 5, ['a', 'b', 'd'], 7]

Question 27

Not sure if it mentioned above or not, but it’s very importable to undestand that .copy() create reference to original object. If you change copied object – you change the original object. .deepcopy() creates new object and does real copying of original object to new one. Changing new deepcopied object doesn’t affect original object.

And yes, .deepcopy() copies original object recursively, while .copy() create a reference object to first-level data of original object.

So the copying/referencing difference between .copy() and .deepcopy() is significant.

Question 28

Deep copy is related to nested structures. If you have list of lists, then deepcopy copies the nested lists also, so it is a recursive copy. With just copy, you have a new outer list, but inner lists are references. Assignment does not copy. For Ex

import copy
spam = [[0, 1, 2, 3], 4, 5]
cheese = copy.copy(spam)
cheese.append(3)
cheese[0].append(3)
print(spam)
print(cheese)

OutPut

[[0, 1, 2, 3, 3], 4, 5] [[0, 1, 2, 3, 3], 4, 5, 3] Copy method copy content of outer list to new list but inner list is still same for both list so if you make changes in inner list of any lists it will affects both list.

But if you use Deep copy then it will create new instance for inner list too.

import copy
spam = [[0, 1, 2, 3], 4, 5]
cheese = copy.deepcopy(spam)
cheese.append(3)
cheese[0].append(3)
print(spam)
print(cheese)

Output

[0, 1, 2, 3] [[0, 1, 2, 3, 3], 4, 5, 3]

Question 29

>>lst=[1,2,3,4,5]

>>a=lst

>>b=lst[:]

>>> b
[1, 2, 3, 4, 5]

>>> a
[1, 2, 3, 4, 5]

>>> lst is b
False

>>> lst is a
True

>>> id(lst)
46263192

>>> id(a)
46263192 ------>  See here id of a and id of lst is same so its called deep copy and even boolean answer is true

>>> id(b)
46263512 ------>  See here id of b and id of lst is not same so its called shallow copy and even boolean answer is false although output looks same.

Question 30

我想知道如何在Jinja中使用另一个变量设置变量。我会解释，我有一个子菜单，我想显示哪个链接处于活动状态。我尝试了这个：

{% set active_link = {{recordtype}} -%}

其中recordtype是为我的模板提供的变量。

Question 31

I would like to know how can I set a variable with another variable in jinja. I will explain, I have got a submenu and I would like show which link is active. I tried this:

{% set active_link = {{recordtype}} -%}

where recordtype is a variable given for my template.

Question 32

{{ }}告诉模板打印值，这在您尝试执行的表达式中将不起作用。而是使用{% set %}template标记，然后以与普通python代码相同的方式分配值。

{% set testing = 'it worked' %}
{% set another = testing %}
{{ another }}

结果：

it worked

Question 33

{{ }} tells the template to print the value, this won’t work in expressions like you’re trying to do. Instead, use the {% set %} template tag and then assign the value the same way you would in normal python code.

{% set testing = 'it worked' %}
{% set another = testing %}
{{ another }}

Result:

it worked

Question 34

多个变量分配的不错简写

{% set label_cls, field_cls = "col-md-7", "col-md-3" %}

Question 35

Nice shorthand for Multiple variable assignments

{% set label_cls, field_cls = "col-md-7", "col-md-3" %}

Question 36

像这样设置

{% set active_link = recordtype -%}

Question 37

Just Set it up like this

{% set active_link = recordtype -%}

Question 38

Is it possible to declare a variable in Python, like so?:

var

so that it initialized to None? It seems like Python allows this, but as soon as you access it, it crashes. Is this possible? If not, why?

EDIT: I want to do this for cases like this:

value

for index in sequence:

   if value == None and conditionMet:
       value = index
       break

Duplicate

Uninitialised value in python (by same author)
Are there any declaration keywords in Python? (by the same author)

This is nice, because you’ll never end up with an uninitialized variable. But be careful — this doesn’t mean that you won’t end up with incorrectly initialized variables. If you init something to None, make sure that’s what you really want, and assign something more meaningful if you can.

Question 40

I’d heartily recommend that you read Other languages have “variables” (I added it as a related link) – in two minutes you’ll know that Python has “names”, not “variables”.

val = None
# ...
if val is None:
   val = any_object

Question 41

In Python 3.6+ you could use Variable Annotations for this:

https://www.python.org/dev/peps/pep-0526/#abstract

PEP 484 introduced type hints, a.k.a. type annotations. While its main focus was function annotations, it also introduced the notion of type comments to annotate variables:

# 'captain' is a string (Note: initial value is a problem)
captain = ...  # type: str

PEP 526 aims at adding syntax to Python for annotating the types of variables (including class variables and instance variables), instead of expressing them through comments:

captain: str  # Note: no initial value!

It seems to be more directly in line with what you were asking “Is it possible only to declare a variable without assigning any value in Python?”

Question 42

I’m not sure what you’re trying to do. Python is a very dynamic language; you don’t usually need to declare variables until you’re actually going to assign to or use them. I think what you want to do is just

foo = None

which will assign the value None to the variable foo.

EDIT: What you really seem to want to do is just this:

#note how I don't do *anything* with value here
#we can just start using it right inside the loop

for index in sequence:
   if conditionMet:
       value = index
       break

try:
    doSomething(value)
except NameError:
    print "Didn't find anything"

It’s a little difficult to tell if that’s really the right style to use from such a short code example, but it is a more “Pythonic” way to work.

EDIT: below is comment by JFS (posted here to show the code)

Unrelated to the OP’s question but the above code can be rewritten as:

for item in sequence:
    if some_condition(item): 
       found = True
       break
else: # no break or len(sequence) == 0
    found = False

if found:
   do_something(item)

NOTE: if some_condition() raises an exception then found is unbound.
NOTE: if len(sequence) == 0 then item is unbound.

The above code is not advisable. Its purpose is to illustrate how local variables work, namely whether “variable” is “defined” could be determined only at runtime in this case. Preferable way:

for item in sequence:
    if some_condition(item):
       do_something(item)
       break

Or

found = False
for item in sequence:
    if some_condition(item):
       found = True
       break

if found:
   do_something(item)

Question 43

I usually initialize the variable to something that denotes the type like

var = ""

or

var = 0

If it is going to be an object then don’t initialize it until you instantiate it:

var = Var()

Question 44

Well, if you want to check if a variable is defined or not then why not check if its in the locals() or globals() arrays? Your code rewritten:

for index in sequence:
   if 'value' not in globals() and conditionMet:
       value = index
       break

If it’s a local variable you are looking for then replace globals() with locals().

Question 45

First of all, my response to the question you’ve originally asked

Q: How do I discover if a variable is defined at a point in my code?

A: Read up in the source file until you see a line where that variable is defined.

But further, you’ve given a code example that there are various permutations of that are quite pythonic. You’re after a way to scan a sequence for elements that match a condition, so here are some solutions:

def findFirstMatch(sequence):
    for value in sequence:
        if matchCondition(value):
            return value

    raise LookupError("Could not find match in sequence")

Clearly in this example you could replace the raise with a return None depending on what you wanted to achieve.

If you wanted everything that matched the condition you could do this:

def findAllMatches(sequence):
    matches = []
    for value in sequence:
        if matchCondition(value):
            matches.append(value)

    return matches

There is another way of doing this with yield that I won’t bother showing you, because it’s quite complicated in the way that it works.

Further, there is a one line way of achieving this:

all_matches = [value for value in sequence if matchCondition(value)]

Question 46

If I’m understanding your example right, you don’t need to refer to ‘value’ in the if statement anyway. You’re breaking out of the loop as soon as it could be set to anything.

value = None
for index in sequence:
   doSomethingHere
   if conditionMet:
       value = index
       break

Question 47

You look like you’re trying to write C in Python. If you want to find something in a sequence, Python has builtin functions to do that, like

value = sequence.index(blarg)

Question 48

It is a good question and unfortunately bad answers as var = None is already assigning a value, and if your script runs multiple times it is overwritten with None every time.

It is not the same as defining without assignment. I am still trying to figure out how to bypass this issue.

Question 49

Is it possible to declare a variable in Python (var=None):

def decl_var(var=None):
if var is None:
    var = []
var.append(1)
return var

Question 50

var_str = str()
var_int = int()

Question 51

If None is a valid data value then you need to the variable another way. You could use:

var = object()

This sentinel is suggested by Nick Coghlan.

Question 52

You can trick an interpreter with this ugly oneliner if None: var = None It do nothing else but adding a variable var to local variable dictionary, not initializing it. Interpreter will throw the UnboundLocalError exception if you try to use this variable in a function afterwards. This would works for very ancient python versions too. Not simple, nor beautiful, but don’t expect much from python.

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问题：在Jinja中设置变量

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重复

有关

Duplicate

Related

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与OP的问题无关，但以上代码可以重写为：

Unrelated to the OP’s question but the above code can be rewritten as:

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有趣好用的Python教程