查找和替换列表中的元素

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.

>>> 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]

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]

List comprehension works well, and looping through with enumerate can save you some memory (b/c the operation’s essentially being done 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]

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]

>>> 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]
>>> 

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

To replace easily all 1 with 10 in a = [1,2,3,4,5,1,2,3,4,5,1]one could use the following one-line lambda+map combination, and ‘Look, Ma, no IFs or FORs!’ :

# This substitutes all '1' with '10' in list 'a' and places result in list 'c':

c = list(map(lambda b: b.replace("1","10"), a))

The following is a very straightforward method in Python 3.x

 a = [1,2,3,4,5,1,2,3,4,5,1]        #Replacing every 1 with 10
 for i in range(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.

I know this is a very old question and there’s a myriad of ways to do it. The simpler one I found is using numpy package.

import numpy

arr = numpy.asarray([1, 6, 1, 9, 8])
arr[ arr == 8 ] = 0 # change all occurrences of 8 by 0
print(arr)

My usecase was replacing None with some default value.

I’ve timed approaches to this problem that were presented here, including the one by @kxr – using str.count.

Test code in ipython with Python 3.8.1:

def rep1(lst, replacer = 0):
    ''' List comprehension, new list '''

    return [item if item is not None else replacer for item in lst]


def rep2(lst, replacer = 0):
    ''' List comprehension, in-place '''    
    lst[:] =  [item if item is not None else replacer for item in lst]

    return lst


def rep3(lst, replacer = 0):
    ''' enumerate() with comparison - in-place '''
    for idx, item in enumerate(lst):
        if item is None:
            lst[idx] = replacer

    return lst


def rep4(lst, replacer = 0):
    ''' Using str.index + Exception, in-place '''

    idx = -1
    # none_amount = lst.count(None)
    while True:
        try:
            idx = lst.index(None, idx+1)
        except ValueError:
            break
        else:
            lst[idx] = replacer

    return lst


def rep5(lst, replacer = 0):
    ''' Using str.index + str.count, in-place '''

    idx = -1
    for _ in range(lst.count(None)):
        idx = lst.index(None, idx+1)
        lst[idx] = replacer

    return lst


def rep6(lst, replacer = 0):
    ''' Using map, return map iterator '''

    return map(lambda item: item if item is not None else replacer, lst)


def rep7(lst, replacer = 0):
    ''' Using map, return new list '''

    return list(map(lambda item: item if item is not None else replacer, lst))


lst = [5]*10**6
# lst = [None]*10**6

%timeit rep1(lst)    
%timeit rep2(lst)    
%timeit rep3(lst)    
%timeit rep4(lst)    
%timeit rep5(lst)    
%timeit rep6(lst)    
%timeit rep7(lst)    

I get:

26.3 ms ± 163 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
29.3 ms ± 206 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
33.8 ms ± 191 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
11.9 ms ± 37.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
11.9 ms ± 60.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
260 ns ± 1.84 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
56.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Using the internal str.index is in fact faster than any manual comparison.

I didn’t know if the exception in test 4 would be more laborious than using str.count, the difference seems negligible.

Note that map() (test 6) returns an iterator and not an actual list, thus test 7.

On long lists and rare occurrences its about 3x faster using list.index() – compared to single step iteration methods presented in the other answers.

def list_replace(lst, old=1, new=10):
    """replace list elements (inplace)"""
    i = -1
    try:
        while 1:
            i = lst.index(old, i + 1)
            lst[i] = new
    except ValueError:
        pass

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]

Find & replace just one item

your_list = [1,2,1]     # replace the first 1 with 11

loc = your_list.index(1)
your_list.remove(1)
your_list.insert(loc, 11)