CODEWITHSUNDEEP
pythonpython-3.xlist
user9987132
user9987132

Reputation:

Changing specific index's in list

Say I had a list:

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]

Would there be any way too take out the first 4 values of the list, make them :

[4, 2, 2, 4]

and the put them back in the list so the list now looks like

[4, 2, 2, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

and do all of this without making the list

[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

at the start.

Also if possible, can you make a loop to do this for all the 4 values available.

So that the list now looks like:

[4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Upvotes: 2

Views: 86

Answers (7)

user9308467
user9308467

Reputation: 1

Here is your original list:

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]

and you'd like to replace the first 4 items with: [4, 2, 2, 4]

Let's assign the new values a new variable. Let's call it update. So:

update = [4, 2, 2, 4]

Here is your code:

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]
update = [4, 2, 2, 4]

for i in range(0, len(update)):
     lis[i] = update[i]

print(lis)

Upvotes: 0

John Luscombe
John Luscombe

Reputation: 383

Given the list of numbers you want to replace the first part of lis with:

replace_lis = [4, 2, 2, 4]

You can iterate over the length of replace_lis, and fill in the index spots in lis accordingly:

for i in range(len(replace_lis)):
    if i < len(lis):
        lis[i] = replace_lis[i]

The i < len(lis) is if replace_lis could be bigger than lis. If there is no chance of that, you can safely omit that if statement.

Upvotes: 0

Vasilis G.
Vasilis G.

Reputation: 7846

Another solution, would be to do the following:

For your first question:

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]

ins = [4, 2, 2, 4]

lis[:len(ins)] = ins

print(lis)

Output:

[4, 2, 2, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

For your second question:

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]

lis = ins * int(len(lis) / len(ins))

print(lis)

Output:

[4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Upvotes: 0

H.Bukhari
H.Bukhari

Reputation: 2251

def list_function(original_list, list_of_value_to_subst):
    original_list[:len(list_of_value_to_subst)]=list_of_value_to_subst

    print (original_list)

    subst_list = list(map(int, list_of_value_to_subst * (len(original_list) // len(list_of_value_to_subst))))

    print (subst_list)


lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]
list_of_value_to_subst=[4, 2, 2, 4]

list_function(lis, list_of_value_to_subst)

output:

[4, 2, 2, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Upvotes: 0

Austin
Austin

Reputation: 26039

Use itertools.cycle and itertools.islice:

from itertools import cycle, islice

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]

out = list(map(int, islice(cycle('4224'), len(lis))))
print(out)

# [4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Or a simpler way (without any imports):

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]

out = list(map(int, '4224' * (len(lis) // 4)))
print(out)

# [4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Upvotes: 1

Ali Shahrivarian
Ali Shahrivarian

Reputation: 311

Just do like this:

lis1 = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]
# change first 4 index
lis1[:4] = 4,2,2,4
print(lis1)
# [4, 2, 2, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]    

lis2 = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]
# change every 4 index in loop
for i in range(0,len(lis2),4):
    lis2[i:i+4] = 4,2,2,4
print(lis2)
# [4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Upvotes: 1

blhsing
blhsing

Reputation: 106445

lis = [0, 0, 0, 0, 0, 0, 0, 0, 0 ,0, 0, 0, 0, 0, 0, 0]
fill = [4, 2, 2, 4]
for i in range(0, len(lis), len(fill)):
    lis[i : i + len(fill)] = fill
print(lis)

This outputs:

[4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 4]

Upvotes: 1

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