结合FOR循环和IF语句的Python方法

问题:结合FOR循环和IF语句的Python方法

我知道如何在单独的行上同时使用for循环和if语句,例如:

>>> a = [2,3,4,5,6,7,8,9,0]
... xyz = [0,12,4,6,242,7,9]
... for x in xyz:
...     if x in a:
...         print(x)
0,4,6,7,9

而且我知道当语句很简单时,我可以使用列表推导来组合这些内容,例如:

print([x for x in xyz if x in a])

但是,我找不到任何地方(复制和学习)的好例子,展示了一组复杂的命令(不仅是“ print x”),这些命令是在for循环和某些if语句组合后发生的。我期望的是:

for x in xyz if x not in a:
    print(x...)

难道这不是python应该工作的方式吗?

I know how to use both for loops and if statements on separate lines, such as:

>>> a = [2,3,4,5,6,7,8,9,0]
... xyz = [0,12,4,6,242,7,9]
... for x in xyz:
...     if x in a:
...         print(x)
0,4,6,7,9

And I know I can use a list comprehension to combine these when the statements are simple, such as:

print([x for x in xyz if x in a])

But what I can’t find is a good example anywhere (to copy and learn from) demonstrating a complex set of commands (not just “print x”) that occur following a combination of a for loop and some if statements. Something that I would expect looks like:

for x in xyz if x not in a:
    print(x...)

Is this just not the way python is supposed to work?


回答 0

您可以使用以下生成器表达式

gen = (x for x in xyz if x not in a)

for x in gen:
    print x

You can use generator expressions like this:

gen = (x for x in xyz if x not in a)

for x in gen:
    print x

回答 1

按照《 Python的禅宗》(如果您想知道代码是否是“ Pythonic”,那就去吧):

  • 美丽胜于丑陋。
  • 显式胜于隐式。
  • 简单胜于复杂。
  • 扁平比嵌套更好。
  • 可读性很重要。

获得两个s 的Pythonic方法是:sorted intersectionset

>>> sorted(set(a).intersection(xyz))
[0, 4, 6, 7, 9]

或那些xyz不在中的元素a

>>> sorted(set(xyz).difference(a))
[12, 242]

但是对于更复杂的循环,您可能希望通过迭代名称良好的生成器表达式和/或调出名称良好的函数来使其扁平化。试图将所有内容都放在一条线上很少是“ Pythonic”的。


在对您的问题和已接受的答案进行其他评论后进行更新

我不确定您要使用的是什么enumerate,但是如果a是字典,则可能要使用这些键,如下所示:

>>> a = {
...     2: 'Turtle Doves',
...     3: 'French Hens',
...     4: 'Colly Birds',
...     5: 'Gold Rings',
...     6: 'Geese-a-Laying',
...     7: 'Swans-a-Swimming',
...     8: 'Maids-a-Milking',
...     9: 'Ladies Dancing',
...     0: 'Camel Books',
... }
>>>
>>> xyz = [0, 12, 4, 6, 242, 7, 9]
>>>
>>> known_things = sorted(set(a.iterkeys()).intersection(xyz))
>>> unknown_things = sorted(set(xyz).difference(a.iterkeys()))
>>>
>>> for thing in known_things:
...     print 'I know about', a[thing]
...
I know about Camel Books
I know about Colly Birds
I know about Geese-a-Laying
I know about Swans-a-Swimming
I know about Ladies Dancing
>>> print '...but...'
...but...
>>>
>>> for thing in unknown_things:
...     print "I don't know what happened on the {0}th day of Christmas".format(thing)
...
I don't know what happened on the 12th day of Christmas
I don't know what happened on the 242th day of Christmas

As per The Zen of Python (if you are wondering whether your code is “Pythonic”, that’s the place to go):

  • Beautiful is better than ugly.
  • Explicit is better than implicit.
  • Simple is better than complex.
  • Flat is better than nested.
  • Readability counts.

The Pythonic way of getting the sorted intersection of two sets is:

>>> sorted(set(a).intersection(xyz))
[0, 4, 6, 7, 9]

Or those elements that are xyz but not in a:

>>> sorted(set(xyz).difference(a))
[12, 242]

But for a more complicated loop you may want to flatten it by iterating over a well-named generator expression and/or calling out to a well-named function. Trying to fit everything on one line is rarely “Pythonic”.


Update following additional comments on your question and the accepted answer

I’m not sure what you are trying to do with enumerate, but if a is a dictionary, you probably want to use the keys, like this:

>>> a = {
...     2: 'Turtle Doves',
...     3: 'French Hens',
...     4: 'Colly Birds',
...     5: 'Gold Rings',
...     6: 'Geese-a-Laying',
...     7: 'Swans-a-Swimming',
...     8: 'Maids-a-Milking',
...     9: 'Ladies Dancing',
...     0: 'Camel Books',
... }
>>>
>>> xyz = [0, 12, 4, 6, 242, 7, 9]
>>>
>>> known_things = sorted(set(a.iterkeys()).intersection(xyz))
>>> unknown_things = sorted(set(xyz).difference(a.iterkeys()))
>>>
>>> for thing in known_things:
...     print 'I know about', a[thing]
...
I know about Camel Books
I know about Colly Birds
I know about Geese-a-Laying
I know about Swans-a-Swimming
I know about Ladies Dancing
>>> print '...but...'
...but...
>>>
>>> for thing in unknown_things:
...     print "I don't know what happened on the {0}th day of Christmas".format(thing)
...
I don't know what happened on the 12th day of Christmas
I don't know what happened on the 242th day of Christmas

回答 2

我个人认为这是最漂亮的版本:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
for x in filter(lambda w: w in a, xyz):
  print x

编辑

如果您非常想避免使用lambda,则可以使用部分函数应用程序并使用运算符模块(该模块提供大多数运算符的功能)。

https://docs.python.org/2/library/operator.html#module-operator

from operator import contains
from functools import partial
print(list(filter(partial(contains, a), xyz)))

I personally think this is the prettiest version:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
for x in filter(lambda w: w in a, xyz):
  print x

Edit

if you are very keen on avoiding to use lambda you can use partial function application and use the operator module (that provides functions of most operators).

https://docs.python.org/2/library/operator.html#module-operator

from operator import contains
from functools import partial
print(list(filter(partial(contains, a), xyz)))

回答 3

以下是接受的答案的一种简化/一种解释:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]

for x in (x for x in xyz if x not in a):
    print(x)

12
242

请注意,generator保持内联。已在python2.7python3.6 (请注意print;)中的括号中对此进行了测试)

The following is a simplification/one liner from the accepted answer:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]

for x in (x for x in xyz if x not in a):
    print(x)

12
242

Notice that the generator was kept inline. This was tested on python2.7 and python3.6 (notice the parens in the print ;) )


回答 4

我可能会使用:

for x in xyz: 
    if x not in a:
        print x...

I would probably use:

for x in xyz: 
    if x not in a:
        print x...

回答 5

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]  
set(a) & set(xyz)  
set([0, 9, 4, 6, 7])
a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]  
set(a) & set(xyz)  
set([0, 9, 4, 6, 7])

回答 6

如果生成器表达式变得过于复杂或复杂,您也可以使用生成器:

def gen():
    for x in xyz:
        if x in a:
            yield x

for x in gen():
    print x

You can use generators too, if generator expressions become too involved or complex:

def gen():
    for x in xyz:
        if x in a:
            yield x

for x in gen():
    print x

回答 7

使用intersectionintersection_update

  • 交叉点

    a = [2,3,4,5,6,7,8,9,0]
    xyz = [0,12,4,6,242,7,9]
    ans = sorted(set(a).intersection(set(xyz)))
  • junction_update

    a = [2,3,4,5,6,7,8,9,0]
    xyz = [0,12,4,6,242,7,9]
    b = set(a)
    b.intersection_update(xyz)

    b是你的答案

Use intersection or intersection_update

  • intersection :

    a = [2,3,4,5,6,7,8,9,0]
    xyz = [0,12,4,6,242,7,9]
    ans = sorted(set(a).intersection(set(xyz)))
    
  • intersection_update:

    a = [2,3,4,5,6,7,8,9,0]
    xyz = [0,12,4,6,242,7,9]
    b = set(a)
    b.intersection_update(xyz)
    

    then b is your answer


回答 8

我喜欢Alex的答案,因为过滤器恰好是应用于列表的if,所以如果您想探索给定条件的列表子集,这似乎是最自然的方法

mylist = [1,2,3,4,5]
another_list = [2,3,4]

wanted = lambda x:x in another_list

for x in filter(wanted, mylist):
    print(x)

此方法对于分离关注点很有用,如果条件函数发生变化,则唯一需要摆弄的代码就是函数本身

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

for x in filter(wanted, mylist):
    print(x)

当您不希望列表的成员时,使用generator方法似乎更好,但是可以对所述成员进行修改,这似乎更适合生成器

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.5 for x in mylist if wanted(x))

for x in generator:
    print(x)

此外,过滤器可与生成器一起使用,尽管在这种情况下效率不高

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.9 for x in mylist)

for x in filter(wanted, generator):
    print(x)

但是,当然,这样写仍然会很好:

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

# for x in filter(wanted, mylist):
for x in mylist if wanted(x):
    print(x)

I liked Alex’s answer, because a filter is exactly an if applied to a list, so if you want to explore a subset of a list given a condition, this seems to be the most natural way

mylist = [1,2,3,4,5]
another_list = [2,3,4]

wanted = lambda x:x in another_list

for x in filter(wanted, mylist):
    print(x)

this method is useful for the separation of concerns, if the condition function changes, the only code to fiddle with is the function itself

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

for x in filter(wanted, mylist):
    print(x)

The generator method seems better when you don’t want members of the list, but a modification of said members, which seems more fit to a generator

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.5 for x in mylist if wanted(x))

for x in generator:
    print(x)

Also, filters work with generators, although in this case it isn’t efficient

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.9 for x in mylist)

for x in filter(wanted, generator):
    print(x)

But of course, it would still be nice to write like this:

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

# for x in filter(wanted, mylist):
for x in mylist if wanted(x):
    print(x)

回答 9

查找列表a和b的唯一公共元素的简单方法:

a = [1,2,3]
b = [3,6,2]
for both in set(a) & set(b):
    print(both)

A simple way to find unique common elements of lists a and b:

a = [1,2,3]
b = [3,6,2]
for both in set(a) & set(b):
    print(both)