Replace Values in List Using Python

Replace values in list using Python

Build a new list with a list comprehension:

new_items = [x if x % 2 else None for x in items]

You can modify the original list in-place if you want, but it doesn't actually save time:

items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for index, item in enumerate(items):
    if not (item % 2):
        items[index] = None

Here are (Python 3.6.3) timings demonstrating the non-timesave:

In [1]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: for index, item in enumerate(items):
   ...:     if not (item % 2):
   ...:         items[index] = None
   ...:
1.06 µs ± 33.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [2]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: new_items = [x if x % 2 else None for x in items]
   ...:
891 ns ± 13.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

And Python 2.7.6 timings:

In [1]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: for index, item in enumerate(items):
   ...:     if not (item % 2):
   ...:         items[index] = None
   ...: 
1000000 loops, best of 3: 1.27 µs per loop
In [2]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: new_items = [x if x % 2 else None for x in items]
   ...: 
1000000 loops, best of 3: 1.14 µs per loop

Finding and replacing elements in a list

You can use the built-in enumerate to get both index and value while iterating the list. Then, use the value to test for a condition and the index to replace that value in the original list:

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

How to replace values of a list with values of another list based on index

Use __getitem__ with map which is quite faster.

list(map(d.__getitem__, a))

Output >> ['d', 'c', 'i', 'h', 'a']

Find and replace string values in list

words = [w.replace('[br]', '<br />') for w in words]

These are called List Comprehensions.

Replace elements in lists based on indexes python

Try this:

rest=[5, 7, 11, 4]
b=[21, 22, 33, 31, 23, 15, 19, 13, 6]
last=[33, 19, 40, 21, 31, 22, 6, 15, 13, 23]

for i, l in enumerate(last):
    if l in b:
        if b.index(l) < len(rest):
            last[i] = rest[b.index(l)]
            
print(last)

Replace values in a list based on condition of previous value

Starting in python 3.8 you now have the walrus operator := which can assign values as part of an expression and works in list comprehensions. You just need to decide what the first value will be if the list starts with 0 since there is no previous value:

alist = [1, 1, 2, 1, 0, 0, 1, 0, 2, 0, 0]

j = 0
[j:= i if i else j for i in alist]
# [1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2]

---

Related Topics