比较Python中两个列表中的值

In Python 2.7, I have two lists of integers:

在Python 2.7中，我有两个整数列表：

x = [1, 3, 2, 0, 2]
y = [1, 2, 2, 3, 1]

I want to create a third list which indicates whether each element in x and y is identical, to yield:

我想创建一个第三个列表，指示x和y中的每个元素是否相同，以产生：

z = [1, 0, 1, 0, 0]

How can I do this using list comprehension?

我怎么能用列表理解来做到这一点？

My attempt is:

我的尝试是：

z = [i == j for i,j in ...]

But I don't know how to complete it.

但我不知道如何完成它。

3 个解决方案

#1

You are looking for zip

你正在寻找拉链

z = [i == j for i,j in zip(x,y)]

But you better add int call to get your desired output

但是你最好添加int调用以获得所需的输出

>>> z = [int(i == j) for i,j in zip(x,y)]
>>> z
[1, 0, 1, 0, 0]

else you'll get a list like [True, False, True, False, False]

否则你会得到一个像[真，假，真，假，假]的清单

As ajcr mentions in a comment, it is better to use itertools.izip instead of zip if the lists are very long. This is because it returns an iterator instead of a list. This is mentioned in the documentation

正如ajcr在评论中提到的，如果列表很长，最好使用itertools.izip而不是zip。这是因为它返回迭代器而不是列表。这在文档中提到

Like zip() except that it returns an iterator instead of a list.

像zip（）一样，除了它返回迭代器而不是列表。

demo

演示

>>> from itertools import izip
>>> z = [int(i == j) for i,j in izip(x,y)]
>>> z
[1, 0, 1, 0, 0]

#2

You can change it a little bit and do:

您可以稍微改变它并执行：

[x[i] == y[i] for i in xrange(len(x))]

If you use Python3 - change xrange to range

如果您使用Python3 - 将xrange更改为范围

#3

While a list comprehension was specified in the question and the answers above are probably better, I thought I'd chime in with a recursive solution:

虽然在问题中指定了列表理解，并且上面的答案可能更好，但我认为我会使用递归解决方案：

def compare_lists(a, b, res=[]):
    if len(a) == len(b):
        if a == []:
            return res
        else:
            if a[0] == b[0]:
                res.append(1)
            else:
                res.append(0)
            return compare_lists(a[1:], b[1:])
    else:
        return "Lists are of different length."

#1