finding and replacing elements in a list (python)

问题:

I have to search through a list and replace all occurrences of one element with another. So far my attempts in code are getting me nowhere, what is the best way to do this?

For example, suppose my list has the following integers

>>> a = [1,2,3,4,5,1,2,3,4,5,1]

and I need to replace all occurrences of the number 1 with the value 10 so the output I need is

>>> a = [10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

Thus my goal is to replace all instances of the number 1 with the number 10.

回答1:

>>> a= [1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1]
>>> for n, i in enumerate(a):
...   if i == 1:
...      a[n] = 10
...
>>> a
[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

回答2:

Try using a list comprehension and the ternary operator.

>>> a=[1,2,3,1,3,2,1,1]
>>> [4 if x==1 else x for x in a]
[4, 2, 3, 4, 3, 2, 4, 4]

回答3:

List comprehension works well--and looping through with enumerate can save you some memory (b/c the operation\'s essentially be doing in place).

There\'s also functional programming...see usage of map:

    >>> a = [1,2,3,2,3,4,3,5,6,6,5,4,5,4,3,4,3,2,1]
    >>> map(lambda x:x if x!= 4 else \'sss\',a)
    [1, 2, 3, 2, 3, \'sss\', 3, 5, 6, 6, 5, \'sss\', 5, \'sss\', 3, \'sss\', 3, 2, 1]

回答4:

If you have several values to replace, you can also use a dictionary:

a = [1, 2, 3, 4, 1, 5, 3, 2, 6, 1, 1]
dic = {1:10, 2:20, 3:\'foo\'}

print([dic.get(n, n) for n in a])

> [10, 20, \'foo\', 4, 10, 5, \'foo\', 20, 6, 10, 10]

回答5:

>>> a=[1,2,3,4,5,1,2,3,4,5,1]
>>> item_to_replace = 1
>>> replacement_value = 6
>>> indices_to_replace = [i for i,x in enumerate(a) if x==item_to_replace]
>>> indices_to_replace
[0, 5, 10]
>>> for i in indices_to_replace:
...     a[i] = replacement_value
... 
>>> a
[6, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6]
>>> 

回答6:

a = [1,2,3,4,5,1,2,3,4,5,1,12]
for i in range (len(a)):
    if a[i]==2:
        a[i]=123

You can use a for and or while loop; however if u know the builtin Enumerate function, then it is recommended to use Enumerate.1

回答7:

The following is a very direct method in Python 2.x

 a = [1,2,3,4,5,1,2,3,4,5,1]        #Replacing every 1 with 10
 for i in xrange(len(a)):
   if a[i] == 1:
     a[i] = 10  
 print a

This method works. Comments are welcome. Hope it helps :)

Also try understanding how outis\'s and damzam\'s solutions work. List compressions and lambda function are useful tools.

回答8:

You can simply use list comprehension in python:

def replace_element(YOUR_LIST, set_to=NEW_VALUE):
    return [i
            if SOME_CONDITION
            else NEW_VALUE
            for i in YOUR_LIST]

for your case, where you want to replace all occurrences of 1 with 10, the code snippet will be like this:

def replace_element(YOUR_LIST, set_to=10):
    return [i
            if i != 1  # keeps all elements not equal to one
            else set_to  # replaces 1 with 10
            for i in YOUR_LIST]