不使用字典将列表中的项目替换为另一个列表中的项目 [英] Replacing an item in list with items of another list without using dictionaries

问题描述

我正在用python开发函数.这是我的内容:

I am developing a function in python. Here is my content:

list = ['cow','orange','mango']
to_replace = 'orange'
replace_with = ['banana','cream']

所以我希望替换后我的列表变成这样

So I want that my list becomes like this after replacement

list = ['cow','banana','cream','mango']

我正在使用此功能:

def replace_item(list, to_replace, replace_with):
    for n,i in enumerate(list):
      if i== to_replace:
         list[n]=replace_with
    return list

它输出如下列表:

['cow', ['banana', 'cream'], 'mango']

那么我如何修改此函数以获取以下输出?

So how do I modify this function to get the below output?

list = ['cow','banana','cream','mango']

注意:我在这里找到了答案:用内容替换列表项另一个列表但是我不想让字典参与其中.我也只想修改我当前的功能，并保持其简单明了.

NOTE: I found an answer here: Replacing list item with contents of another list but I don't want to involve dictionaries in this. I also want to modify my current function only and keep it simple and straight forward.

推荐答案

首先，切勿使用python内置名称和关键字作为变量名称(将 list 更改为 ls ).

First off, never use python built-in names and keywords as your variable names (change the list to ls).

您不需要循环，找到索引然后链接切片:

You don't need loop, find the index then chain the slices:

In [107]: from itertools import chain    
In [108]: ls = ['cow','orange','mango']
In [109]: to_replace = 'orange'
In [110]: replace_with = ['banana','cream']
In [112]: idx = ls.index(to_replace)  
In [116]: list(chain(ls[:idx], replace_with, ls[idx+1:]))
Out[116]: ['cow', 'banana', 'cream', 'mango']

在python 3.5+中，您可以使用就地解压缩:

In python 3.5+ you can use in-place unpacking:

[*ls[:idx], *replace_with, *ls[idx+1:]]

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