Python 实用宝典

Question 1

Assume that S and T are assigned sets. Without using the join operator |, how can I find the union of the two sets? This, for example, finds the intersection:

S = {1, 2, 3, 4}
T = {3, 4, 5, 6}
S_intersect_T = { i for i in S if i in T }

So how can I find the union of two sets in one line without using |?

Question 2

You can use union method for sets: set.union(other_set)

Note that it returns a new set i.e it doesn’t modify itself.

Question 3

You could use or_ alias:

>>> from operator import or_
>>> from functools import reduce # python3 required
>>> reduce(or_, [{1, 2, 3, 4}, {3, 4, 5, 6}])
set([1, 2, 3, 4, 5, 6])

Question 4

If you are fine with modifying the original set (which you may want to do in some cases), you can use set.update():

S.update(T)

The return value is None, but S will be updated to be the union of the original S and T.

Question 5

Assuming you also can’t use s.union(t), which is equivalent to s | t, you could try

>>> from itertools import chain
>>> set(chain(s,t))
set([1, 2, 3, 4, 5, 6])

Or, if you want a comprehension,

>>> {i for j in (s,t) for i in j}
set([1, 2, 3, 4, 5, 6])

Question 6

If by join you mean union, try this:

set(list(s) + list(t))

It’s a bit of a hack, but I can’t think of a better one liner to do it.

Question 7

Suppose you have 2 lists

 A = [1,2,3,4]
 B = [3,4,5,6]

so you can find A Union B as follow

 union = set(A).union(set(B))

also if you want to find intersection and non-intersection you do that as follow

 intersection = set(A).intersection(set(B))
 non_intersection = union - intersection

Question 8

You can just unpack both sets into one like this:

>>> set_1 = {1, 2, 3, 4}
>>> set_2 = {3, 4, 5, 6}
>>> union = {*set_1, *set_2}
>>> union
{1, 2, 3, 4, 5, 6}

The * unpacks the set. Unpacking is where an iterable (e.g. a set or list) is represented as every item it yields. This means the above example simplifies to {1, 2, 3, 4, 3, 4, 5, 6} which then simplifies to {1, 2, 3, 4, 5, 6} because the set can only contain unique items.

Question 9

You can do union or simple list comprehension

[A.add(_) for _ in B]

A would have all the elements of B

Question 10

I’m trying to make a set of sets in Python. I can’t figure out how to do it.

Starting with the empty set xx:

xx = set([])
# Now we have some other set, for example
elements = set([2,3,4])
xx.add(elements)

but I get

TypeError: unhashable type: 'list'

or

TypeError: unhashable type: 'set'

Is it possible to have a set of sets in Python?

I am dealing with a large collection of sets and I want to be able to not have to deal duplicate sets (a set B of sets A1, A2, …., An would “cancel” two sets if Ai = Aj)

Question 11

Python’s complaining because the inner set objects are mutable and thus not hashable. The solution is to use frozenset for the inner sets, to indicate that you have no intention of modifying them.

Question 12

People already mentioned that you can do this with a frozenset(), so I will just add a code how to achieve this:

For example you want to create a set of sets from the following list of lists:

t = [[], [1, 2], [5], [1, 2, 5], [1, 2, 3, 4], [1, 2, 3, 6]]

you can create your set in the following way:

t1 = set(frozenset(i) for i in t)

Question 13

Use frozenset inside.

Question 14

So I had the exact same problem. I wanted to make a data structure that works as a set of sets. The problem is that the sets must contain immutable objects. So, what you can do is simply make it as a set of tuples. That worked fine for me!

A = set()
A.add( (2,3,4) )##adds the element
A.add( (2,3,4) )##does not add the same element
A.add( (2,3,5) )##adds the element, because it is different!

Question 15

As of 2020, the official Python documentation advise using frozenset to represent sets of sets.

Question 16

Recently I noticed that when I am converting a list to set the order of elements is changed and is sorted by character.

Consider this example:

x=[1,2,20,6,210]
print x 
# [1, 2, 20, 6, 210] # the order is same as initial order

set(x)
# set([1, 2, 20, 210, 6]) # in the set(x) output order is sorted

My questions are –

Why is this happening?
How can I do set operations (especially Set Difference) without losing the initial order?

Question 17

A set is an unordered data structure, so it does not preserve the insertion order.
This depends on your requirements. If you have an normal list, and want to remove some set of elements while preserving the order of the list, you can do this with a list comprehension:
```
>>> a = [1, 2, 20, 6, 210]
>>> b = set([6, 20, 1])
>>> [x for x in a if x not in b]
[2, 210]
```
If you need a data structure that supports both fast membership tests and preservation of insertion order, you can use the keys of a Python dictionary, which starting from Python 3.7 is guaranteed to preserve the insertion order:
```
>>> a = dict.fromkeys([1, 2, 20, 6, 210])
>>> b = dict.fromkeys([6, 20, 1])
>>> dict.fromkeys(x for x in a if x not in b)
{2: None, 210: None}
```
b doesn’t really need to be ordered here – you could use a set as well. Note that a.keys() - b.keys() returns the set difference as a set, so it won’t preserve the insertion order.

In older versions of Python, you can use collections.OrderedDict instead:
```
>>> a = collections.OrderedDict.fromkeys([1, 2, 20, 6, 210])
>>> b = collections.OrderedDict.fromkeys([6, 20, 1])
>>> collections.OrderedDict.fromkeys(x for x in a if x not in b)
OrderedDict([(2, None), (210, None)])
```

Question 18

~~In Python 3.6, set() now should keep the order, but~~ there is another solution for Python 2 and 3:

>>> x = [1, 2, 20, 6, 210]
>>> sorted(set(x), key=x.index)
[1, 2, 20, 6, 210]

Question 19

Answering your first question, a set is a data structure optimized for set operations. Like a mathematical set, it does not enforce or maintain any particular order of the elements. The abstract concept of a set does not enforce order, so the implementation is not required to. When you create a set from a list, Python has the liberty to change the order of the elements for the needs of the internal implementation it uses for a set, which is able to perform set operations efficiently.

Question 20

remove duplicates and preserve order by below function

def unique(sequence):
    seen = set()
    return [x for x in sequence if not (x in seen or seen.add(x))]

check this link

Question 21

In mathematics, there are sets and ordered sets (osets).

set: an unordered container of unique elements (Implemented)
oset: an ordered container of unique elements (NotImplemented)

In Python, only sets are directly implemented. We can emulate osets with regular dict keys (3.7+).

Given

a = [1, 2, 20, 6, 210, 2, 1]
b = {2, 6}

Code

oset = dict.fromkeys(a).keys()
# dict_keys([1, 2, 20, 6, 210])

Demo

Replicates are removed, insertion-order is preserved.

list(oset)
# [1, 2, 20, 6, 210]

Set-like operations on dict keys.

oset - b
# {1, 20, 210}

oset | b
# {1, 2, 5, 6, 20, 210}

oset & b
# {2, 6}

oset ^ b
# {1, 5, 20, 210}

Details

Note: an unordered structure does not preclude ordered elements. Rather, maintained order is not guaranteed. Example:

assert {1, 2, 3} == {2, 3, 1}                    # sets (order is ignored)

assert [1, 2, 3] != [2, 3, 1]                    # lists (order is guaranteed)

One may be pleased to discover that a list and multiset (mset) are two more fascinating, mathematical data structures:

list: an ordered container of elements that permits replicates (Implemented)
mset: an unordered container of elements that permits replicates (NotImplemented)*

Summary

Container | Ordered | Unique | Implemented
----------|---------|--------|------------
set       |    n    |    y   |     y
oset      |    y    |    y   |     n
list      |    y    |    n   |     y
mset      |    n    |    n   |     n*

^{*A multiset can be indirectly emulated with collections.Counter(), a dict-like mapping of multiplicities (counts).}

Question 22

As denoted in other answers, sets are data structures (and mathematical concepts) that do not preserve the element order –

However, by using a combination of sets and dictionaries, it is possible that you can achieve wathever you want – try using these snippets:

# save the element order in a dict:
x_dict = dict(x,y for y, x in enumerate(my_list) )
x_set = set(my_list)
#perform desired set operations
...
#retrieve ordered list from the set:
new_list = [None] * len(new_set)
for element in new_set:
   new_list[x_dict[element]] = element

Question 23

Building on Sven’s answer, I found using collections.OrderedDict like so helped me accomplish what you want plus allow me to add more items to the dict:

import collections

x=[1,2,20,6,210]
z=collections.OrderedDict.fromkeys(x)
z
OrderedDict([(1, None), (2, None), (20, None), (6, None), (210, None)])

If you want to add items but still treat it like a set you can just do:

z['nextitem']=None

And you can perform an operation like z.keys() on the dict and get the set:

z.keys()
[1, 2, 20, 6, 210]

Question 24

An implementation of the highest score concept above that brings it back to a list:

def SetOfListInOrder(incominglist):
    from collections import OrderedDict
    outtemp = OrderedDict()
    for item in incominglist:
        outtemp[item] = None
    return(list(outtemp))

Tested (briefly) on Python 3.6 and Python 2.7.

Question 25

In case you have a small number of elements in your two initial lists on which you want to do set difference operation, instead of using collections.OrderedDict which complicates the implementation and makes it less readable, you can use:

# initial lists on which you want to do set difference
>>> nums = [1,2,2,3,3,4,4,5]
>>> evens = [2,4,4,6]
>>> evens_set = set(evens)
>>> result = []
>>> for n in nums:
...   if not n in evens_set and not n in result:
...     result.append(n)
... 
>>> result
[1, 3, 5]

Its time complexity is not that good but it is neat and easy to read.

Question 26

It’s interesting that people always use ‘real world problem’ to make joke on the definition in theoretical science.

If set has order, you first need to figure out the following problems. If your list has duplicate elements, what should the order be when you turn it into a set? What is the order if we union two sets? What is the order if we intersect two sets with different order on the same elements?

Plus, set is much faster in searching for a particular key which is very good in sets operation (and that’s why you need a set, but not list).

If you really care about the index, just keep it as a list. If you still want to do set operation on the elements in many lists, the simplest way is creating a dictionary for each list with the same keys in the set along with a value of list containing all the index of the key in the original list.

def indx_dic(l):
    dic = {}
    for i in range(len(l)):
        if l[i] in dic:
            dic.get(l[i]).append(i)
        else:
            dic[l[i]] = [i]
    return(dic)

a = [1,2,3,4,5,1,3,2]
set_a  = set(a)
dic_a = indx_dic(a)

print(dic_a)
# {1: [0, 5], 2: [1, 7], 3: [2, 6], 4: [3], 5: [4]}
print(set_a)
# {1, 2, 3, 4, 5}

Question 27

Here’s an easy way to do it:

x=[1,2,20,6,210]
print sorted(set(x))

Question 28

I’m learning python, and I have a novice question about initializing sets. Through testing, I’ve discovered that a set can be initialized like so:

my_set = {'foo', 'bar', 'baz'}

Are there any disadvantages of doing it this way, as opposed to the standard way of:

my_set = set(['foo', 'bar', 'baz'])

or is it just a question of style?

Question 29

There are two obvious issues with the set literal syntax:

my_set = {'foo', 'bar', 'baz'}

It’s not available before Python 2.7
There’s no way to express an empty set using that syntax (using {} creates an empty dict)

Those may or may not be important to you.

The section of the docs outlining this syntax is here.

Question 30

Compare also the difference between {} and set() with a single word argument.

>>> a = set('aardvark')
>>> a
{'d', 'v', 'a', 'r', 'k'} 
>>> b = {'aardvark'}
>>> b
{'aardvark'}

but both a and b are sets of course.

Question 31

From Python 3 documentation (the same holds for python 2.7):

Curly braces or the set() function can be used to create sets. Note: to create an empty set you have to use set(), not {}; the latter creates an empty dictionary, a data structure that we discuss in the next section.

in python 2.7:

>>> my_set = {'foo', 'bar', 'baz', 'baz', 'foo'}
>>> my_set
set(['bar', 'foo', 'baz'])

Be aware that {} is also used for map/dict:

>>> m = {'a':2,3:'d'}
>>> m[3]
'd'
>>> m={}
>>> type(m)
<type 'dict'>

One can also use comprehensive syntax to initialize sets:

>>> a = {x for x in """didn't know about {} and sets """ if x not in 'set' }
>>> a
set(['a', ' ', 'b', 'd', "'", 'i', 'k', 'o', 'n', 'u', 'w', '{', '}'])

Question 32

You need to do empty_set = set() to initialize an empty set. {} is am empty dictionaty.

Question 33

Given Zero Piraeus’ answer to another question, we have that

x = tuple(set([1, "a", "b", "c", "z", "f"]))
y = tuple(set(["a", "b", "c", "z", "f", 1]))
print(x == y)

Prints True about 85% of the time with hash randomization enabled. Why 85%?

Question 34

I’m going to assume any readers of this question to have read both:

The first thing to note is that hash randomization is decided on interpreter start-up.

The hash of each letter will be the same for both sets, so the only thing that can matter is if there is a collision (where order will be affected).

By the deductions of that second link we know the backing array for these sets starts at length 8:

_ _ _ _ _ _ _ _

In the first case, we insert 1:

_ 1 _ _ _ _ _ _

and then insert the rest:

α 1 ? ? ? ? ? ?

Then it is rehashed to size 32:

    1 can't collide with α as α is an even hash
  ↓ so 1 is inserted at slot 1 first
? 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

In the second case, we insert the rest:

? β ? ? ? ? ? ?

And then try to insert 1:

    Try to insert 1 here, but will
  ↓ be rehashed if β exists
? β ? ? ? ? ? ?

And then it will be rehashed:

    Try to insert 1 here, but will
    be rehashed if β exists and has
  ↓ not rehashed somewhere else
? β ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

So whether the iteration orders are different depends solely on whether β exists.

The chance of a β is the chance that any of the 5 letters will hash to 1 modulo 8 and hash to 1 modulo 32.

Since anything that hashes to 1 modulo 32 also hashes to 1 modulo 8, we want to find the chance that of the 32 slots, one of the five is in slot 1:

5 (number of letters) / 32 (number of slots)

5/32 is 0.15625, so there is a 15.625% chance¹ of the orders being different between the two set constructions.

Not very strangely at all, this is exactly what Zero Piraeus measured.

_{¹Technically even this isn’t obvious. We can pretend every one of the 5 hashes uniquely because of rehashing, but because of linear probing it’s actually more likely for “bunched” structures to occur… but because we’re only looking at whether a single slot is occupied, this doesn’t actually affect us.}

Question 35

How come when I change the order of the two sets in the unions below, I get different results?

set1 = {1, 2, 3}
set2 = {True, False}

print(set1 | set2)
# {False, 1, 2, 3}

print(set2 | set1)
#{False, True, 2, 3}

Question 36

Why the union() doesn’t contain all items

The 1 and True are equivalent and considered to be duplicates. Likewise the 0 and False are equivalent as well:

>>> 1 == True
True
>>> 0 == False
True

Which equivalent value is used

When multiple equivalent values are encountered, sets keep the first one seen:

>>> {0, False}
{0}
>>> {False, 0}
{False}

Ways to make the values be distinct

To get them to be treated as distinct, just store them in a (value, type) pair:

>>> set1 = {(1, int), (2, int), (3, int)}
>>> set2 = {(True, bool), (False, bool)}
>>> set1 | set2
{(3, <class 'int'>), (1, <class 'int'>), (2, <class 'int'>),
 (True, <class 'bool'>), (False, <class 'bool'>)}
>>> set1 & set2
set()

Another way to make the values distinct is to store them as strings:

>>> set1 = {'1', '2', '3'}
>>> set2 = {'True', 'False'}
>>> set1 | set2
{'2', '3', 'False', 'True', '1'}
>>> set1 & set2
set()

Hope this clears up the mystery and shows the way forward :-)

Rescued from the comments:

This is the standard technique for breaking cross-type equivalence (i.e. 0.0 == 0, True == 1, and Decimal(8.5) == 8.5). The technique is used in Python 2.7’s regular expression module to force unicode regexes to be cached distinctly from otherwise equivalent str regexes. The technique is also used in Python 3 for functools.lru_cache() when the typed parameter is true.

If the OP needs something other than the default equivalence relation, then some new relation needs to be defined. Depending the use case, that could be case-insensitivity for strings, normalization for unicode, visual appearance (things that look different are considered different), identity (no two distinct objects are considered equal), a value/type pair, or some other function that defines an equivalence relation. Given the OPs specific example, it would seem that he/she expected either distinction by type or visual distinction.

Question 37

In Python, False and 0 are considered equivalent, as are True and 1. Because True and 1 are considered the same value, only one of them can be present in a set a the same time. Which one depends on the order they are added to the set in. In the first line, set1 is used as the first set, so we get 1 in the resulting set. In the second set, True is in the first set, so True is included in the result.

Question 38

If you look at https://docs.python.org/3/library/stdtypes.html#boolean-values section 4.12.10. Boolean Values:

Boolean values are the two constant objects False and True. They are used to represent truth values (although other values can also be considered false or true). In numeric contexts (for example when used as the argument to an arithmetic operator), they behave like the integers 0 and 1, respectively.

Question 39

The comparison operator (==, !=) is defined for boolean True and False to match 1 and 0.

That’s why, in the set union, when it checks whether True is in the new set already, it gets a truthy answer:

>>> True in {1}
True
>>> 1 in {True}
True

Question 40

I have the following procedure:

def myProc(invIndex, keyWord):
    D={}
    for i in range(len(keyWord)):
        if keyWord[i] in invIndex.keys():
                    D.update(invIndex[query[i]])
    return D

But I am getting the following error:

Traceback (most recent call last):
  File "<stdin>", line 3, in <module>
TypeError: cannot convert dictionary update sequence element #0 to a sequence

I do not get any error if D contains elements. But I need D to be empty at the beginning.

Question 41

D = {} is a dictionary not set.

>>> d = {}
>>> type(d)
<type 'dict'>

Use D = set():

>>> d = set()
>>> type(d)
<type 'set'>
>>> d.update({1})
>>> d.add(2)
>>> d.update([3,3,3])
>>> d
set([1, 2, 3])

Question 42

>>> d = {}
>>> D = set()
>>> type(d)
<type 'dict'>
>>> type(D)
<type 'set'>

What you’ve made is a dictionary and not a Set.

The update method in dictionary is used to update the new dictionary from a previous one, like so,

>>> abc = {1: 2}
>>> d.update(abc)
>>> d
{1: 2}

Whereas in sets, it is used to add elements to the set.

>>> D.update([1, 2])
>>> D
set([1, 2])

Question 43

When you assign a variable to empty curly braces {} eg: new_set = {}, it becomes a dictionary. To create an empty set, assign the variable to a ‘set()’ ie: new_set = set()

Question 44

给定一套

{0, 1, 2, 3}

如何产生子集：

[set(),
 {0},
 {1},
 {2},
 {3},
 {0, 1},
 {0, 2},
 {0, 3},
 {1, 2},
 {1, 3},
 {2, 3},
 {0, 1, 2},
 {0, 1, 3},
 {0, 2, 3},
 {1, 2, 3},
 {0, 1, 2, 3}]

Question 45

Given a set

{0, 1, 2, 3}

How can I produce the subsets:

[set(),
 {0},
 {1},
 {2},
 {3},
 {0, 1},
 {0, 2},
 {0, 3},
 {1, 2},
 {1, 3},
 {2, 3},
 {0, 1, 2},
 {0, 1, 3},
 {0, 2, 3},
 {1, 2, 3},
 {0, 1, 2, 3}]

Question 46

Pythonitertools页面对此有一个精确的powerset配方：

from itertools import chain, combinations

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

输出：

>>> list(powerset("abcd"))
[(), ('a',), ('b',), ('c',), ('d',), ('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd'), ('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'd'), ('b', 'c', 'd'), ('a', 'b', 'c', 'd')]

如果您不喜欢开头的空元组，则可以更改range语句range(1, len(s)+1)以避免使用0长度的组合。

Question 47

The Python itertools page has exactly a powerset recipe for this:

from itertools import chain, combinations

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

Output:

>>> list(powerset("abcd"))
[(), ('a',), ('b',), ('c',), ('d',), ('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd'), ('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'd'), ('b', 'c', 'd'), ('a', 'b', 'c', 'd')]

If you don’t like that empty tuple at the beginning, you can just change the range statement to range(1, len(s)+1) to avoid a 0-length combination.

Question 48

这是有关电源组的更多代码。这是从头开始写的：

>>> def powerset(s):
...     x = len(s)
...     for i in range(1 << x):
...         print [s[j] for j in range(x) if (i & (1 << j))]
...
>>> powerset([4,5,6])
[]
[4]
[5]
[4, 5]
[6]
[4, 6]
[5, 6]
[4, 5, 6]

马克·鲁沙科夫（Mark Rushakoff）的评论在这里适用：“如果您不喜欢开头的空元组，请继续。”您可以将range语句更改为range（1，len（s）+1）以避免长度为0的组合”除了在我的情况下更改for i in range(1 << x)到for i in range(1, 1 << x)。

回到今年以后，我现在将其编写为：

def powerset(s):
    x = len(s)
    masks = [1 << i for i in range(x)]
    for i in range(1 << x):
        yield [ss for mask, ss in zip(masks, s) if i & mask]

然后，测试代码如下所示：

print(list(powerset([4, 5, 6])))

使用yield意味着您无需在单个内存中计算所有结果。在主循环之外预先计算掩码被认为是值得进行的优化。

Question 49

Here is more code for a powerset. This is written from scratch:

>>> def powerset(s):
...     x = len(s)
...     for i in range(1 << x):
...         print [s[j] for j in range(x) if (i & (1 << j))]
...
>>> powerset([4,5,6])
[]
[4]
[5]
[4, 5]
[6]
[4, 6]
[5, 6]
[4, 5, 6]

Mark Rushakoff’s comment is applicable here: “If you don’t like that empty tuple at the beginning, on.”you can just change the range statement to range(1, len(s)+1) to avoid a 0-length combination”, except in my case you change for i in range(1 << x) to for i in range(1, 1 << x).

Returning to this years later, I’d now write it like this:

def powerset(s):
    x = len(s)
    masks = [1 << i for i in range(x)]
    for i in range(1 << x):
        yield [ss for mask, ss in zip(masks, s) if i & mask]

And then the test code would look like this, say:

print(list(powerset([4, 5, 6])))

Using yield means that you do not need to calculate all results in a single piece of memory. Precalculating the masks outside the main loop is assumed to be a worthwhile optimization.

Question 50

如果您正在寻找一个快速的答案，我刚刚在Google上搜索了“ python power set”，并提出了以下建议：Python Power Set Generator

这是该页面中代码的复制粘贴：

def powerset(seq):
    """
    Returns all the subsets of this set. This is a generator.
    """
    if len(seq) <= 1:
        yield seq
        yield []
    else:
        for item in powerset(seq[1:]):
            yield [seq[0]]+item
            yield item

可以这样使用：

 l = [1, 2, 3, 4]
 r = [x for x in powerset(l)]

现在r是您想要的所有元素的列表，可以进行排序和打印：

r.sort()
print r
[[], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]]

Question 51

If you’re looking for a quick answer, I just searched “python power set” on google and came up with this: Python Power Set Generator

Here’s a copy-paste from the code in that page:

def powerset(seq):
    """
    Returns all the subsets of this set. This is a generator.
    """
    if len(seq) <= 1:
        yield seq
        yield []
    else:
        for item in powerset(seq[1:]):
            yield [seq[0]]+item
            yield item

This can be used like this:

 l = [1, 2, 3, 4]
 r = [x for x in powerset(l)]

Now r is a list of all the elements you wanted, and can be sorted and printed:

r.sort()
print r
[[], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]]

Question 52

def powerset(lst):
    return reduce(lambda result, x: result + [subset + [x] for subset in result],
                  lst, [[]])

Question 53

def powerset(lst):
    return reduce(lambda result, x: result + [subset + [x] for subset in result],
                  lst, [[]])

Question 54

powerset有一个改进：

def powerset(seq):
    """
    Returns all the subsets of this set. This is a generator.
    """
    if len(seq) <= 0:
        yield []
    else:
        for item in powerset(seq[1:]):
            yield [seq[0]]+item
            yield item

Question 55

There is a refinement of powerset:

def powerset(seq):
    """
    Returns all the subsets of this set. This is a generator.
    """
    if len(seq) <= 0:
        yield []
    else:
        for item in powerset(seq[1:]):
            yield [seq[0]]+item
            yield item

Question 56

TL; DR（直接进入简化）

我知道我以前已经添加了答案，但是我真的很喜欢我的新实现。我将一个集合作为输入，但是实际上它可以是任何迭代的，并且我返回的是集合的集合，即输入的幂集。我喜欢这种方法，因为它更符合幂集（所有子集）的数学定义。

def power_set(A):
    """A is an iterable (list, tuple, set, str, etc)
    returns a set which is the power set of A."""
    length = len(A)
    l = [a for a in A]
    ps = set()

    for i in range(2 ** length):
        selector = f'{i:0{length}b}'
        subset = {l[j] for j, bit in enumerate(selector) if bit == '1'}
        ps.add(frozenset(subset))

    return ps

如果您想确切地在答案中发布输出，请使用以下命令：

>>> [set(s) for s in power_set({1, 2, 3, 4})]
[{3, 4},
 {2},
 {1, 4},
 {2, 3, 4},
 {2, 3},
 {1, 2, 4},
 {1, 2},
 {1, 2, 3},
 {3},
 {2, 4},
 {1},
 {1, 2, 3, 4},
 set(),
 {1, 3},
 {1, 3, 4},
 {4}]

说明

已知功率集的元素数为2 ** len(A)，因此可以在for循环中清楚地看到。

我需要将输入（最好是一组）转换为列表，因为一组是唯一无序元素的数据结构，而顺序对于生成子集至关重要。

selector是此算法的关键。请注意，selector它的长度与输入集的长度相同，为了使之成为可能，它使用带填充的f字符串。基本上，这使我可以选择将在每次迭代期间添加到每个子集的元素。假设输入集包含3个元素{0, 1, 2}，那么选择器将采用0到7（含）之间的值，二进制形式为：

因此，无论是否应添加原始集合的元素，每个位都可以用作指示符。查看二进制数字，然后将每个数字都视为超集的元素，这1意味着j应添加索引处的元素，并且不应添加0此元素。

我使用集合推导在每次迭代时生成一个子集，并将此子集转换为，frozenset以便可以将其添加到ps（幂集）。否则，我将无法添加它，因为Python中的集合仅包含不可变的对象。

简化版

您可以使用一些python理解来简化代码，因此可以摆脱那些for循环。您还zip可以避免使用j索引，并且代码最终将如下所示：

def power_set(A):
    length = len(A)
    return {
        frozenset({e for e, b in zip(A, f'{i:{length}b}') if b == '1'})
        for i in range(2 ** length)
    }

而已。我喜欢这种算法的原因是它比其他算法更清晰，更直观，因为itertools即使它按预期工作，依靠它看起来也很神奇。

Question 57

TL;DR (go directly to Simplification)

I know I have previously added an answer, but I really like my new implementation. I am taking a set as input, but it actually could be any iterable, and I am returning a set of sets which is the power set of the input. I like this approach because it is more aligned with the mathematical definition of power set (set of all subsets).

def power_set(A):
    """A is an iterable (list, tuple, set, str, etc)
    returns a set which is the power set of A."""
    length = len(A)
    l = [a for a in A]
    ps = set()

    for i in range(2 ** length):
        selector = f'{i:0{length}b}'
        subset = {l[j] for j, bit in enumerate(selector) if bit == '1'}
        ps.add(frozenset(subset))

    return ps

If you want exactly the output you posted in your answer use this:

>>> [set(s) for s in power_set({1, 2, 3, 4})]
[{3, 4},
 {2},
 {1, 4},
 {2, 3, 4},
 {2, 3},
 {1, 2, 4},
 {1, 2},
 {1, 2, 3},
 {3},
 {2, 4},
 {1},
 {1, 2, 3, 4},
 set(),
 {1, 3},
 {1, 3, 4},
 {4}]

Explanation

It is known that the number of elements of the power set is 2 ** len(A), so that could clearly be seen in the for loop.

I need to convert the input (ideally a set) into a list because by a set is a data structure of unique unordered elements, and the order will be crucial to generate the subsets.

selector is key in this algorithm. Note that selector has the same length as the input set, and to make this possible it is using an f-string with padding. Basically, this allows me to select the elements that will be added to each subset during each iteration. Let’s say the input set has 3 elements {0, 1, 2}, so selector will take values between 0 and 7 (inclusive), which in binary are:

So, each bit could serve as an indicator if an element of the original set should be added or not. Look at the binary numbers, and just think of each number as an element of the super set in which 1 means that an element at index j should be added, and 0 means that this element should not be added.

I am using a set comprehension to generate a subset at each iteration, and I convert this subset into a frozenset so I can add it to ps (power set). Otherwise, I won’t be able to add it because a set in Python consists only of immutable objects.

Simplification

You can simplify the code using some python comprehensions, so you can get rid of those for loops. You can also use zip to avoid using j index and the code will end up as the following:

def power_set(A):
    length = len(A)
    return {
        frozenset({e for e, b in zip(A, f'{i:{length}b}') if b == '1'})
        for i in range(2 ** length)
    }

That’s it. What I like of this algorithm is that is clearer and more intuitive than others because it looks quite magical to rely on itertools even though it works as expected.

Question 58

def get_power_set(s):
  power_set=[[]]
  for elem in s:
    # iterate over the sub sets so far
    for sub_set in power_set:
      # add a new subset consisting of the subset at hand added elem
      power_set=power_set+[list(sub_set)+[elem]]
  return power_set

例如：

get_power_set([1,2,3])

让

[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

Question 59

def get_power_set(s):
  power_set=[[]]
  for elem in s:
    # iterate over the sub sets so far
    for sub_set in power_set:
      # add a new subset consisting of the subset at hand added elem
      power_set=power_set+[list(sub_set)+[elem]]
  return power_set

For example:

get_power_set([1,2,3])

yield

[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

Question 60

我发现以下算法非常清楚和简单：

def get_powerset(some_list):
    """Returns all subsets of size 0 - len(some_list) for some_list"""
    if len(some_list) == 0:
        return [[]]

    subsets = []
    first_element = some_list[0]
    remaining_list = some_list[1:]
    # Strategy: get all the subsets of remaining_list. For each
    # of those subsets, a full subset list will contain both
    # the original subset as well as a version of the subset
    # that contains first_element
    for partial_subset in get_powerset(remaining_list):
        subsets.append(partial_subset)
        subsets.append(partial_subset[:] + [first_element])

    return subsets

生成功率集的另一种方法是生成所有具有n位的二进制数。作为幂集，带n数字的位数为2 ^ n。该算法的原理是，子集中可能存在或不存在元素，因为二进制数字可能是一个或零，但不能同时存在。

def power_set(items):
    N = len(items)
    # enumerate the 2 ** N possible combinations
    for i in range(2 ** N):
        combo = []
        for j in range(N):
            # test bit jth of integer i
            if (i >> j) % 2 == 1:
                combo.append(items[j])
        yield combo

在上MITx时，我找到了两种算法：6.00.2x计算思维和数据科学概论，我认为这是我所见过的最容易理解的算法之一。

Question 61

I have found the following algorithm very clear and simple:

def get_powerset(some_list):
    """Returns all subsets of size 0 - len(some_list) for some_list"""
    if len(some_list) == 0:
        return [[]]

    subsets = []
    first_element = some_list[0]
    remaining_list = some_list[1:]
    # Strategy: get all the subsets of remaining_list. For each
    # of those subsets, a full subset list will contain both
    # the original subset as well as a version of the subset
    # that contains first_element
    for partial_subset in get_powerset(remaining_list):
        subsets.append(partial_subset)
        subsets.append(partial_subset[:] + [first_element])

    return subsets

Another way one can generate the powerset is by generating all binary numbers that have n bits. As a power set the amount of number with n digits is 2 ^ n. The principle of this algorithm is that an element could be present or not in a subset as a binary digit could be one or zero but not both.

def power_set(items):
    N = len(items)
    # enumerate the 2 ** N possible combinations
    for i in range(2 ** N):
        combo = []
        for j in range(N):
            # test bit jth of integer i
            if (i >> j) % 2 == 1:
                combo.append(items[j])
        yield combo

I found both algorithms when I was taking MITx: 6.00.2x Introduction to Computational Thinking and Data Science, and I consider it is one of the easiest algorithms to understand I have seen.

Question 62

我只是想提供最容易理解的解决方案，即反代码高尔夫版本。

from itertools import combinations

l = ["x", "y", "z", ]

def powerset(items):
    combo = []
    for r in range(len(items) + 1):
        #use a list to coerce a actual list from the combinations generator
        combo.append(list(combinations(items,r)))
    return combo

l_powerset = powerset(l)

for i, item in enumerate(l_powerset):
    print "All sets of length ", i
    print item

结果

全部套长0

[()]

全部套长1

[('x',), ('y',), ('z',)]

全套长度2

[('x', 'y'), ('x', 'z'), ('y', 'z')]

全套长度3

[('x', 'y', 'z')]

有关更多信息，请参见itertools文档，以及有关电源集的Wikipedia条目

Question 63

I just wanted to provide the most comprehensible solution, the anti code-golf version.

from itertools import combinations

l = ["x", "y", "z", ]

def powerset(items):
    combo = []
    for r in range(len(items) + 1):
        #use a list to coerce a actual list from the combinations generator
        combo.append(list(combinations(items,r)))
    return combo

l_powerset = powerset(l)

for i, item in enumerate(l_powerset):
    print "All sets of length ", i
    print item

The results

All sets of length 0

[()]

All sets of length 1

[('x',), ('y',), ('z',)]

All sets of length 2

[('x', 'y'), ('x', 'z'), ('y', 'z')]

All sets of length 3

[('x', 'y', 'z')]

For more see the itertools docs, also the wikipedia entry on power sets

Question 64

只是一个快速的动力设定刷新器！

X的幂集，简单来说就是X的所有子集的集合，包括空集

示例集X =（a，b，c）

幂集= {{a，b，c}，{a，b}，{a，c}，{b，c}，{a}，{b}，{c}，{}}

这是查找功率集的另一种方法：

def power_set(input):
    # returns a list of all subsets of the list a
    if (len(input) == 0):
        return [[]]
    else:
        main_subset = [ ]
        for small_subset in power_set(input[1:]):
            main_subset += [small_subset]
            main_subset += [[input[0]] + small_subset]
        return main_subset

print(power_set([0,1,2,3]))

完全归功于来源

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TL; DR（直接进入简化）

说明

简化版

TL;DR (go directly to Simplification)

Explanation

Simplification

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