CODEWITHSUNDEEP
pythonpython-3.xlist
Pavel.D
Pavel.D

Reputation: 581

Keep growing values in list and hold list size

Few days ago I have been trying many ways and methods for at get desired behaviour of list in Python. I need to construct a method, which assigns and changes values in list at the same time it keeps size of list. let me explain

I have such a code. 

A = [1,1,1,1,1]
B=[]

for i in range(0,len(A)):
    for j in range(1,3):
        val = j*2*A[i]
        B.insert(i,val)
    print(B)

The result I get:

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

What I want to do!

take the first value of A-list in this case int 1. then make some calculation by increasing a range function. val = 1*2*A[0] = 2 and set in the B-list, and continue val = 2*2*A[0] = 4 and set in B-list. then next val = 1*2*A[1] = 2 and val = 2*2*A[1] = 4. I am expecting to have a output like this.

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

Update:

I have updated a sample of code, this time a for-loop is added for construction of list A but it runs into some problem.which I do not really understand why it does run with error list index out of range. That is clear, when len(list A) is 4, len(list B) becomes 5. it is not expected list B gets out of range? 

A = []
B = [] 
for n in range(1,11):
    A.append(n/n)
    for i in range(0,len(A)):
        B.insert(i, A[i])
        for j in range(1,3):
            for k, _ in enumerate(B):
                B[k] = j*2*A[k]
                print(B)

Output I get:

[2.0]
[4.0]
[2.0, 4.0]       #<------- Not need it 
[2.0, 2.0]
[4.0, 2.0]       #<------- Not need it 
[4.0, 4.0]
[2.0, 1.0, 4.0]  #<------- Not need it 
[2.0, 2.0, 4.0]  #<------- Not need it 

IndexError: list index out of range

Expected output.

[2.0]
[4.0]
[2.0, 2.0]
[4.0, 4.0]
---------------

I do not really know how to solve this type of problem, I appreciate any help...

Upvotes: 4

Views: 854

Answers (1)

SpghttCd
SpghttCd

Reputation: 10860

you should put the insertfunction in the outer loop, otherwise you insert not only for every element in A, but also for every calculation step in range(1, 3)

i.e.

A = [1,1,1,1,1]
B=[]

for i in range(0,len(A)):
    B.insert(i, A[i])
    for j in range(1,3):
        B[i] = j*2*A[i]
    print(B)

or better Python:

A = [1, 1, 1, 1, 1]
B = []

for i, elmnt in enumerate(A):
    B.insert(i, elmnt)
    for j in range(1, 3):
        B[i] = j * 2 * elmnt
    print(B)

But why do yo want to write val = 1*2*A[0] = 2 into B[0] if you overwrite it anyway in the next step by val = 2*2*A[0] = 4?
That doesn't make sense.

EDIT:
Latest agreement about what should be the expected behaviour:

A = [1, 1, 1, 1, 1]
B = []

for i, elmnt in enumerate(A):
    B.insert(i, elmnt)
    for j in range(1, 3):
        for k, _ in enumerate(B):
            B[k] = j * 2 * elmnt
        print(B)
[2]
[4]
[2, 2]
[4, 4]
[2, 2, 2]
[4, 4, 4]
[2, 2, 2, 2]
[4, 4, 4, 4]
[2, 2, 2, 2, 2]
[4, 4, 4, 4, 4]

...or with numpy it's easier to change all values of an array at once:

import numpy as np
A = np.array([1, 1, 1, 1, 1], 'int')
B = np.array([], 'int')

for i, elmnt in enumerate(A):
    B = np.append(B, elmnt)
    for j in range(1, 3):
        B[:] = j * 2 * elmnt
        print(B)

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

Upvotes: 3

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