finding and replacing elements in a list

Question

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?<

Answer 1

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]

Answer 2

for i in range(0, len(lst)):
    lst.insert(i, lst[i])
    lst.remove(lst[i+1])

Answer 3

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]

Answer 4

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.

Answer 5

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

Answer 6

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.