Array problems in for loop

Question

I want to go from one array A of 10 elements to the array B of 100 elements.

Each element of B from 0 to 9 is equal to the element 0 of A

Each element of B from 10 to 19 is equal to the element 1 of A

....

Each element of B from 90 to 99 is equal to the element 9 of A

I did the following code but it does not work

a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
A = np.asarray(a)

b = []
for i in range(len(A)*10):
    b.append(0)
    
B = np.asarray(b)   

for i in range(len(A)):
    for j in range(9):
        B[j]=A[i]

Expected result:

B [ 0,0,0,0,0,0,0,0,0,0,
  1,1,1,1,1,1,1,1,1,1,
  2,2,2,2,2,2,2,2,2,2
  ...,
  9,9,9,9,9,9,9,9,9,9 ]

Please format your code

Shivam Jha
– Shivam Jha

2020-07-03 11:43:30 +00:00
Commented Jul 3, 2020 at 11:43 — Shivam Jha
– Shivam Jha, Commented Jul 3, 2020 at 11:43
Have a look at the numpy.repeat function; docs here.

s3dev
– s3dev

2020-07-03 11:47:59 +00:00
Commented Jul 3, 2020 at 11:47 — s3dev
– s3dev, Commented Jul 3, 2020 at 11:47
Best answer thanks!!!

chrisis
– chrisis

2020-07-03 12:35:38 +00:00
Commented Jul 3, 2020 at 12:35 — chrisis
– chrisis, Commented Jul 3, 2020 at 12:35

xszym · Accepted Answer · 2020-07-03 12:32:39Z

2

You are saving values only in first 9 list elements. You have to 'scale' it by adding i*10 to index.

import numpy as np

a=[0, 1, 2, 3, 4, 5, 6, 7]
A = np.asarray(a)

b = []
for i in range(len(A)**2):
    b.append(0)
    
B = np.asarray(b)   

for i in range(len(A)):
    for j in range(len(A)):
        B[j + i*len(A)]=A[i]

print(B)

edited Jul 3, 2020 at 12:32

answered Jul 3, 2020 at 11:49

xszym

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2 Comments

chrisis Over a year ago

Actually by using a nother vector as A I obtain every 9 value a 0. I cannot understand why

xszym Over a year ago

I edited my answer so it can be more scalable. I hope it will work now

lenik · Accepted Answer · 2020-07-03 11:51:42Z

1

This works for me:

>>> a = [1,2,3]
>>> [ x for i in a for x in [i]*3]
[1, 1, 1, 2, 2, 2, 3, 3, 3]
>>>

You may replace 3 with 10 or whatever you like.

Answering the question from Jacob:

>>> [[a]*10 for a in A]
[[1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2], [3, 3, 3, 3, 3, 3, 3, 3, 3, 3]]

edited Jul 3, 2020 at 11:51

answered Jul 3, 2020 at 11:47

lenik

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3 Comments

JakobVinkas Over a year ago

The double for loop seems a bit over kill (harder to understand) when you can instead use [[a]*10 for a in A]

lenik Over a year ago

@JakobVinkas you'll get list of lists, not a simple list from your code. try running it in python

JakobVinkas Over a year ago

Oh you are correct, I wrote my answer before the question was edited to clearly display the structure of the B array, my bad.

Mark · Accepted Answer · 2020-07-03 12:01:57Z

You should avoid loops with numpy whenever possible. It kind of defeats the point. Here you can just use repeat():

import numpy as np

a=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
A = np.asarray(a)
B = A.repeat(10)

B:

array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2,
       2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4,
       4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6,
       6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8,
       8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9])

If want the a nested list, just reshape:

B = A.repeat(10).reshape(-1, 10)

array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
       [3, 3, 3, 3, 3, 3, 3, 3, 3, 3],
       [4, 4, 4, 4, 4, 4, 4, 4, 4, 4],
       [5, 5, 5, 5, 5, 5, 5, 5, 5, 5],
       [6, 6, 6, 6, 6, 6, 6, 6, 6, 6],
       [7, 7, 7, 7, 7, 7, 7, 7, 7, 7],
       [8, 8, 8, 8, 8, 8, 8, 8, 8, 8],
       [9, 9, 9, 9, 9, 9, 9, 9, 9, 9]])

JakobVinkas · Accepted Answer · 2020-07-03 11:46:26Z

0

You can use numpy and specify how many iterations of each element you want:

import numpy as np
A = [1,2,3,4]
B = [np.full(10, a) for a in A]
print(B)

Or if you prefer to not use numpy, instead use:

A = [1,2,3,4]
B = [[a]*10 for a in A]
print(B)

Giving you the wanted list B

answered Jul 3, 2020 at 11:46

JakobVinkas

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2 Comments

JakobVinkas Over a year ago

My answer is not the best one, but in this case a referes to a in A making such that a will take iterating values (all of the values) of A from start to end of the array

chrisis Over a year ago

Ok I understood it is the element of the array. But, now I get the following results: [[1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2], [3, 3, 3, 3, 3, 3, 3, 3, 3, 3], [4, 4, 4, 4, 4, 4, 4, 4, 4, 4]] Is it possible to compact everything in one array? In this way: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]

ThunderPhoenix · Accepted Answer · 2020-07-03 11:53:25Z

0

Try this:

a = [*range(10)]
b = []
for i in range(10):
    b.extend([a[i]* 10])
B = np.asarray(b)

answered Jul 3, 2020 at 11:53

ThunderPhoenix

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Comments

wa007 · Accepted Answer · 2020-07-03 12:13:44Z

0

a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = []
for x in a:
    b += [x] * 10
print b

This answer is better, idea from lenik

a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [x for x in a for i in range(10)]
print b

edited Jul 3, 2020 at 12:13

answered Jul 3, 2020 at 11:50

wa007

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3 Comments

chrisis Over a year ago

Hi, how would be possible to avoid the double square brakets?

wa007 Over a year ago

I don't think it can avoid.

wa007 Over a year ago

The answer [x for i in a for x in [i] * 10] from lenik is nice.

Ashish Karn · Accepted Answer · 2020-07-03 12:25:52Z

0

Answer in a single line: print([item for sublist in [[i]*10 for i in range(1,10)] for item in sublist])

answered Jul 3, 2020 at 12:25

Ashish Karn

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Comments

gboffi · Accepted Answer · 2020-07-04 16:33:15Z

0

If a were a generic list and not an ordered sequence

In [20]: a = [1, 'a', 3.14159, False, {1:2, 3:4}]

you could do as follows

In [21]: [_ for _ in (zip(*(a for _ in a))) for _ in _]
Out[21]: 
[1,
 1,
 1,
 1,
 1,
 'a',
 'a',
 'a',
 'a',
 'a',
 3.14159,
 3.14159,
 3.14159,
 3.14159,
 3.14159,
 False,
 False,
 False,
 False,
 False,
 {1: 2, 3: 4},
 {1: 2, 3: 4},
 {1: 2, 3: 4},
 {1: 2, 3: 4},
 {1: 2, 3: 4}]

edited Jul 4, 2020 at 16:33

answered Jul 3, 2020 at 12:03

gboffi

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Comments

Syscall · Accepted Answer · 2023-02-14 18:17:00Z

-1

list1=[]
list2=[]
for i in range (0,10,1):
    list1.append(i)
print(list1)
for i in range (0,10,1):
    for j in range (0,10,1):
        j=i
        list2.append(j)
print(list2)

edited Feb 14, 2023 at 18:17

Syscall

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answered Feb 14, 2023 at 17:26

Nur Sayed

1

1 Comment

Syscall Over a year ago

Please add some explanations.

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