有没有办法按索引合并多个列表索引？ [英] Is there a way to merge multiple list index by index?

问题描述

例如，我有三个（相同长度）列表

For example, I have three lists (of the same length)

A = [1,2,3]
B = [a,b,c]
C = [x,y,z]

我希望将它合并为：
[[1，a，x]，[2，b，y]，[3，c，z]]。

and i want to merge it into something like: [[1,a,x],[2,b,y],[3,c,z]].

以下是我目前的情况：

define merger(A,B,C):
  answer = 
  for y in range (len(A)):
    a = A[y]
    b = B[y]
    c = C[y]
    temp = [a,b,c]
    answer = answer.extend(temp)
  return answer

收到错误：

'NoneType'对象没有属性'extend'

'NoneType' object has no attribute 'extend'

推荐答案

看起来您的代码应该说 answer = [] ，而将其删除会导致问题。但你遇到的主要问题是：

It looks like your code is meant to say answer = [], and leaving that out will cause problems. But the major problem you have is this:

answer = answer.extend(temp)

extend 修改回答 并返回None。将其保留为 answer.extend（temp），它将起作用。您可能还想使用追加方法而不是 extend - 追加puts 一个对象 （列表 temp ）在 answer 结尾处，而 extend 单独追加每个项目的temp，最终给出你所追求的扁平化版本： [1，'a'，'x'，2，'b' ，'y'，3，'c'，'z'] 。

extend modifies answer and returns None. Leave this as just answer.extend(temp) and it will work. You likely also want to use the append method rather than extend - append puts one object (the list temp) at the end of answer, while extend appends each item of temp individually, ultimately giving the flattened version of what you're after: [1, 'a', 'x', 2, 'b', 'y', 3, 'c', 'z'].

但是，不是重新发明轮子，这正是内置的 zip 适用于：

But, rather than reinventing the wheel, this is exactly what the builtin zip is for:

>>> A = [1,2,3]
>>> B = ['a', 'b', 'c']
>>> C = ['x', 'y', 'z']
>>> list(zip(A, B, C))
[(1, 'a', 'x'), (2, 'b', 'y'), (3, 'c', 'z')]

请注意，在Python 2中， zip 返回元组列表;在Python 3中，它返回一个惰性迭代器（即，它根据请求构建元组，而不是预先计算它们）。如果你想要Python 3中的Python 2行为，你将通过 list 传递它，就像我上面所做的那样。如果您想要Python 2中的Python 3行为，请使用 <$ c来自itertools的$ c> izip 。

Note that in Python 2, zip returns a list of tuples; in Python 3, it returns a lazy iterator (ie, it builds the tuples as they're requested, rather than precomputing them). If you want the Python 2 behaviour in Python 3, you pass it through list as I've done above. If you want the Python 3 behaviour in Python 2, use the function izip from itertools.

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