How to replace elements in a nested list python 3?

Question

How to replace elements in a particular position nested list python 3? i want to insert elements in a[1] position. The inserted elements are in list b.

a = [['04\01\1997','alphanum4569874','22','4.0'],['07\01\1997','Anee_69213654','23','2.0']]

b = ['alphanum1','alphanum2']

for idx,item in enumerate(b):
    for i in b:
        a[1].append(i)

print(a)

Expected Output:

[['04\01\1997','alphanum1','22','4.0'['07\01\1997','alphanum1','23','2.0']]

Current Output:

[['04\x01\x01997', 'alphanum4569874', '22', '4.0'], ['07\x01\x01997', 'Anee_69213654', '23', '2.0', 'alphanum1', 'alphanum2', 'alphanum1', 'alphanum2']]

I don't think that's the expected output, since the number of brackets doesn't match - did you mean [['04\01\1997','alphanum1','22','4.0'], ['07\01\1997','alphanum2','23','2.0']]? So, basically, you're looking to replace the 2nd element of the sublists of the first list with elements of the second list? — Grismar
– Grismar, Commented Jun 17, 2019 at 3:36

Grismar · Accepted Answer · 2019-06-17 03:52:39Z

2

I'm assuming your list b will always have the same number of elements as there are lists in a and that you always want to replace (not insert) the second element.

This works:

for ax, bx in zip(a, b):
    ax[1] = bx

First zip pairs up every element from a with the corresponding element from b, the for loop then gives you each pair one at a time as ax and bx. Since ax is a list, it's actually the list in a (not a copy) and with a[1] = bx you're just overwriting the second element in the list with bx.

Another way of doing it would be with a list comprehension:

new_a = [[ax[0], bx, *ax[2:]] for ax, bx in zip(a, b)]

This has the advantage of not modifying the original a, but gives you a new list with the replacements.

What happens here is that it still uses zip for the pairing, but instead of replacing ax[1], it makes a new list with the first element of ax, followed by bx and then followed by the rest of ax from the third element onwards. Notice the * in there - that 'explodes' the list into its separate elements, so they can be added to the new list.

You also asked about avoiding zip, although I think this is a worse solution:

for i in range(len(a)):
    a[i][1] = b[i]

What this does is let i run from 0 to the length of a minus one, replacing the 2nd element of each element of a one at a time with the matching element from b, by using i to index both a and b.

edited Jun 17, 2019 at 3:52

answered Jun 17, 2019 at 3:44

Grismar

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2 Comments

suba Over a year ago

thank ypu so much. i don't know the usage of zip. so any alternative for this solution other than using zip? i want solution in simplify

Grismar Over a year ago

I'll add an example not using zip, but from my perspective it would be more complicated than zip - zip is a very useful function to know, since you don't have to keep track of your position in both the lists.

Ajax1234 · Accepted Answer · 2019-06-17 03:33:45Z

2

You can use zip:

a = [['04\01\1997','alphanum4569874','22','4.0'],['07\01\1997','Anee_69213654','23','2.0']]
b = ['alphanum1','alphanum2']
result = [[c, d, *j] for d, [c, _, *j] in zip(b, a)]

Output:

[['04\x01\x01997', 'alphanum1', '22', '4.0'], ['07\x01\x01997', 'alphanum2', '23', '2.0']]

answered Jun 17, 2019 at 3:33

Ajax1234

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Comments

itsvinayak · Accepted Answer · 2019-06-17 04:08:04Z

1

this code works

a = [['04\01\1997','alphanum4569874','22','4.0'],['07\01\1997','Anee_69213654','23','2.0']]

b = ['alphanum1','alphanum2']
c=0
for i in a:
    i[1]=b[c]
    c+=1
print(a)

output

[['04\x01\x01997', 'alphanum1', '22', '4.0'], ['07\x01\x01997', 'alphanum2', '23', '2.0']]

edited Jun 17, 2019 at 4:08

answered Jun 17, 2019 at 3:41

itsvinayak

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3 Comments

suba Over a year ago

thank you so much but its not replacing its just appending

itsvinayak Over a year ago

in question it was give to insert this is my output [['04\x01\x01997', 'alphanum1', 'alphanum4569874', '22', '4.0'], ['07\x01\x01997', 'alphanum2', 'Anee_69213654', '23', '2.0']]

suba Over a year ago

i want [['04\x01\x01997', 'alphanum1', '22', '4.0'], ['07\x01\x01997', 'alphanum2', '23', '2.0']]

U13-Forward · Accepted Answer · 2019-06-17 03:36:47Z

0

Or try:

print([[x[0], y, *x[2:]] for x, y in zip(a, b)])

Output:

[['04\x01\x01997', 'alphanum1', '22', '4.0'], ['07\x01\x01997', 'alphanum2', '23', '2.0']]

answered Jun 17, 2019 at 3:36

U13-Forward

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2 Comments

suba Over a year ago

thank ypu so much. i don't know the usage of zip. so any alternative for this solution other than using zip? i want solution in simplify

U13-Forward Over a year ago

@suba You should learn zip instead, here is the documentation: python-reference.readthedocs.io/en/latest/docs/functions/…

drum · Accepted Answer · 2019-06-17 03:31:53Z

-1

I think what you want is this:

 for item in a:
    for i in b:
       item[1].append(i)

 print (a)

answered Jun 17, 2019 at 3:31

drum

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1 Comment

suba Over a year ago

'str' object has no attribute 'append' i m getting this erro how to resolve

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