Python：查找特定列表项并替换

Question

I tried to look for the solution to this question by perusing the forum but many of the explanations provided are too advanced for me and was wondering if there were a simpler method.

So my question is:

my_list = [1, 2, 3, 4, 5, 6]

In my_list , I would like to replace all even numbers with an "x".

Is there method where I can just go through a loop and replace them?

Ok, so I was hoping there'd be a simple method like a my_list.replace function but realized there wasn't. So here is the context in which I need to locate and replace:

def censor(text, word):
    string = text.split()
    for n in range(len(string)):
        if string[n] == word:
            string[n] == "*" * len(word)

    return string

print censor("hi bob how are you bob", "bob")

So in this program I'm trying to replace the "bob" in the phrase to asterisks in accordance to the length of the word bob. But I am not sure what I'm doing wrong here. Instead the output I'm getting is the same phrase. Thanks!

Answer 1

Use a list comprehension:

my_list = ['x' if i % 2 == 0 else i for i in my_list]

Yes, this still uses a loop, but you need to scan all values for the even numbers anyway .

Answer 2

If you don't want a list comprehension (if you wanted to have a more elaborate replacement at some point later) you could loop through it and edit each value in turn if necessary:

for i in range(len(my_list)):
    if not my_list[i] % 2:
        my_list[i] = 'x'

It's not as clean as a comprehension though.

Answer 3

To "just go through the loop and replace them", you could use:

for index, item in enumerate(my_list):
    if not item % 2:
        my_list[index] = "x"

A list comprehension is neater, but will build a new list rather than modifying the old one - which you need will depend on what you are doing.

Answer 4

为了提供另一种选择，我们还可以使用map ：

map(lambda n: 'x' if n%2==0 else n, my_list)

Answer 5

This is the most concise:

map(lambda i: 'x' if i % 2 ==0 else i, my_list)

Out[17]: [1, 'x', 3, 'x', 5, 'x']