Multiplying each sublist by a constant

Question

l2=[[13,11,9],[7,5,3],[1]]

I want to multiply each sublist in list l2 with a constant number i.e 13*1,11*1,9*1 and 7*2,5*2,3*2 and 1*3 and the final result would be 13,11,9 and 14,10,6 and 3.

All that has been posted is a program description. However, we need you to [ask a question](//stackoverflow.com/help/how-to-ask). We can't be sure what you want from us. Please [edit] your post to include a valid question that we can answer. Reminder: make sure you know [what is on-topic here](//stackoverflow.com/help/on-topic); **asking us to write the program for you**, suggestions, and external links are **off-topic.** — Patrick Artner, Feb 11 '19 at 18:24
This is a more general versionof https://stackoverflow.com/questions/43671999/python-multiply-each-item-in-sublists-by-a-list .. if the list would contain the same number instead of different ones... similarly question is here https://stackoverflow.com/questions/26446338/how-to-multiply-all-integers-inside-list/26446368 and there are about a dozend more on SO already — Patrick Artner, Feb 11 '19 at 18:26

score 1 · Answer 1 · answered Feb 11 '19 at 18:29

1

Using enumerate

Ex:

l2=[[13,11,9],[7,5,3],[1]]
print([[j*i for j in v] for i, v in enumerate(l2, 1)])

Output:

[[13, 11, 9], [14, 10, 6], [3]]

answered Feb 11 '19 at 18:29

Rakesh

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score 0 · Answer 2 · answered Feb 11 '19 at 18:29

0

You can leverage enumerate here:

l2 = [[13,11,9],[7,5,3],[1]]

l3 = [[x * (idx + 1) for x in sublist] for idx, sublist in enumerate(l2)]
print(l3)

This yields

[[13, 11, 9], [14, 10, 6], [3]]

answered Feb 11 '19 at 18:29

Jan

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Type Errror:'int' object is not iterable – Swaraj Kotapati Feb 11 '19 at 18:43
Your error makes no sense. At least not with this code. – Jan Feb 11 '19 at 19:22

score 0 · Answer 3 · answered Feb 11 '19 at 18:30

0

Try this:

l2=[[13,11,9],[7,5,3],[1]]

counter = 1
for i in range(len(l2)):
    for j in range(len(l2[i])):
        l2[i][j] *= counter        
    counter += 1

print(l2)

answered Feb 11 '19 at 18:30

Heyran.rs

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score 0 · Answer 4 · answered Feb 11 '19 at 18:30

0

Is it always the first sublist is 1, the 2nd is two, etc?

for i in range(0, len(l2)):
    i += 1
    print(f'{l2[i-1]} times {i}')
    print(list(map(lambda x: x*i, l2[i-1])))

answered Feb 11 '19 at 18:30

Patrick Conwell

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Yes the first sublist is always 1 – Swaraj Kotapati Feb 11 '19 at 18:38
1

why not `range(1,len(l2)+1)` to remove an extra step – mad_ Feb 11 '19 at 18:41
I thought about that, too. Figured it was about 50/50 on which was better. – Patrick Conwell Feb 11 '19 at 23:00

mad_ · Answer 5 · 2019-02-11T18:48:27.667

0

One liner with map and enumerate

[map((idx).__mul__, sublist) for idx,sublist in enumerate(l2,1)] #[[13, 11, 9], [14, 10, 6], [3]]

using numpy.multiply

import numpy as np
[np.multiply(sublist,[idx]).tolist() for idx,sublist in enumerate(l2,1)]

edited Feb 11 '19 at 18:48

answered Feb 11 '19 at 18:32

mad_

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ncica · Answer 6 · 2019-02-12T09:06:16.553

new_list = []
multiplier = 1 
# passing over sublists( e.g. [13,11,9] = li) in l2
for li in l2: 
    # multiply all elements in sublist(li) with multiplier(in first step multiplier = 1)
    # append the multiplied sublist to the new list
    new_list.append([li[i]*multiplier for i in range(len(li))])
    # multiplier increase by one (e.g. in second step we need to multiply elements from second sublist with multiplier = 2)
    multiplier += 1
print (new_list)

Multiplying each sublist by a constant

6 Answers6