Python 偶数一个列表？
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Even numbers a list?
提问by
How do I create a list and only extract or search out the even numbers in that list?
如何创建列表并仅提取或搜索该列表中的偶数？
Create a function even_only(l)that takes a list of integers as its only argument. The
function will return a new list containing all (and only) the elements of l which are evenly divisible by 2. The original list l shall remain unchanged.
创建一个even_only(l)将整数列表作为其唯一参数的函数。该函数将返回一个包含所有（且仅）l 中可被 2 整除的元素的新列表。原始列表 l 应保持不变。
For examples, even_only([1, 3, 6, 10, 15, 21, 28])should return [6, 10, 28], and
even_only([1, 4, 9, 16, 25])should return [4, 16].
例如，even_only([1, 3, 6, 10, 15, 21, 28])应该返回[6, 10, 28]，并且
even_only([1, 4, 9, 16, 25])应该返回[4, 16]。
Hint: Start by creating an empty list, and whenever you encounter an even number in it, add it to your list, then at the end, return your list.
提示：首先创建一个空列表，每当您遇到偶数时，将其添加到您的列表中，然后在最后返回您的列表。
采纳答案by Mark Rushakoff
"By hand":
“用手”：
def even_only(lst):
evens = []
for number in lst:
if is_even(number):
evens.append(number)
return evens
Pythonic:
蟒蛇：
def even_only(iter):
return [x for x in iter if is_even(x)]
Since it's homework, you can fill in the is_evenfunction.
既然是作业，就可以填is_even函数了。
回答by Donut
Simplest way would be to do what you posted in a comment -- iterate through the input list to find digits evenly divisible by 2, and add them to the return list if so.
最简单的方法是执行您在评论中发布的操作 - 遍历输入列表以查找可被 2 整除的数字，如果是，则将它们添加到返回列表中。
The list.append(x)function will help you add an item to a list.
该list.append(x)功能将帮助您将项目添加到列表中。
Also as mentioned, look at using the modulooperation to determine if a number is divisible by 2...
同样如上所述，看看使用模运算来确定一个数字是否可以被 2 整除......
回答by Vincent Savard
The best way to do this (as a beginner) is probably a comprehension list. Since this is a homework, I won't do it for you, but here is the syntax :
做到这一点的最好方法（作为初学者）可能是一个理解列表。由于这是一个家庭作业，我不会为你做，但这里是语法：
[x for x in your_list if (your condition)]
You just have to replace (your condition) with what fits well (basically, exactly what you described).
您只需将（您的状况）替换为合适的（基本上，正是您所描述的）。
P.S. I know some people may say comprehension lists are a bit advanced for a beginner, but I think it is not a concept too hard to catch and extremely useful.
PS 我知道有些人可能会说理解列表对于初学者来说有点高级，但我认为它不是一个很难掌握的概念，而且非常有用。
回答by Troy J. Farrell
Use the filterfunction to do this in a functional way:
使用该filter函数以函数方式执行此操作：
>>> even_filter = lambda x: not x % 2
>>> result = filter(even_filter, [0, 1, 2, 3, 4])
>>> assert result == [0, 2, 4]
Edit: updated with the correct parity of zero per Vincent's comment.
编辑：根据文森特的评论更新了正确的零奇偶校验。
回答by u478344
>>> a = [1, 3, 6, 10, 15, 21, 28]
>>> b = [i for i in a if i%2 ==0 ]
>>> b
[6, 10, 28]
>>> a
[1, 3, 6, 10, 15, 21, 28]
回答by PaulMcG
>>> even_only = lambda seq : [ x for x in seq if str(x)[-1] in "02468" ]
>>> even_only([1, 3, 6, 10, 15, 21, 28])
[6, 10, 28]
回答by Vlox
I recently had this issue and used:
我最近遇到了这个问题并使用了：
list=[1,2,3,4,5,6] #whatever your list is, just a sample
evens=[x for x in list if np.mod(x,2)==0]
print evens
returns [2,4,6]
返回 [2,4,6]
回答by Yuriy Samorodov
All = (1, 3, 6, 10, 15, 21, 28] From itertools import takewhile Evenonly = takewhile(lambda x: x%2 == 0, All) Print (list (Evenonly)
All = (1, 3, 6, 10, 15, 21, 28] from itertools import takewhile Evenonly = takewhile(lambda x: x%2 == 0, All) Print (list (Evenonly)
回答by Vivek Singh
Seems a bit late.But whats given below, works fine for me:
似乎有点晚了。但是下面给出的内容对我来说很好用：
def even_list(*args):
# Returns the even numbers in the list
return [x for x in args if (x % 2 == 0)]
even_list(1,2,3,4,5,6,7,8)
[2, 4, 6, 8]
回答by Ioannis Nasios
for every element in list, if element of list modulo is 0 then number is even
对于列表中的每个元素，如果列表中的元素取模为 0，则数字为偶数
initial_list = [1,22,13,41,15,16,87]
even_list = [ x for x in initial_list if x % 2 == 0]
even_list
# [22, 16]
