替换列表python列表中的元素

本文介绍了替换列表python列表中的元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个列表列表，如下所示:

I have a list of lists as follows:

list=[] *some code to append elements to list* list=[['a','bob'],['a','bob'],['a','john']]

我要浏览此列表，并将'bob'的所有实例更改为'b'，而其他实例保持不变.

I want to go through this list and change all instances of 'bob to 'b' and leave others unchanged.

for x in list: for a in x: if "bob" in a: a.replace("bob", 'b')

在打印出x后，它仍然与列表相同，但不如下:

After printing out x it is still the same as list, but not as follows:

list=[['a','b'],['a','b'],['a','john']]

为什么更改未反映在列表中?

Why is the change not being reflected in list?

推荐答案

由于str.replace无法就地运行 ，因此返回副本.作为不可变对象，您需要将字符串分配到列表列表中的元素.

Because str.replace doesn't work in-place, it returns a copy. As immutable objects, you need to assign the strings to elements in your list of lists.

如果通过enumerate提取索引整数，则可以直接分配给列表列表:

You can assign directly to your list of lists if you extract indexing integers via enumerate:

L = [['a','bob'],['a','bob'],['a','john']] for i, x in enumerate(L): for j, a in enumerate(x): if 'bob' in a: L[i][j] = a.replace('bob', 'b')

结果:

[['a', 'b'], ['a', 'b'], ['a', 'john']]

更多Pythonic将使用列表理解来创建新列表.例如，如果两个值中只有第二个包含需要检查的名称:

More Pythonic would be to use a list comprehension to create a new list. For example, if only the second of two values contains names which need checking:

L = [[i, j if j != 'bob' else 'b'] for i, j in L]