Convert for loop outcome to numpy array

Question

I want to create an array with 17 elements starting with 1 and other numbers are each twice the value immediately before it.

what I have so far is:

import numpy as np


array = np.zeros(shape=17)

array[0]=1

x = 1

for i in array:
    print(x)
    x *= 2

print(array)

what I got is:

1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536
[1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]

and what I want is:

[1.2.4.8.16.32.64.128.256.512.1024.2048.4096.8192.16384.32768.65536]

Paul Panzer · Accepted Answer · 2019-09-18 09:49:26Z

1

There is a function for that

np.logspace(0,16,17,base=2,dtype=int)
# array([    1,     2,     4,     8,    16,    32,    64,   128,   256,
#          512,  1024,  2048,  4096,  8192, 16384, 32768, 65536])

Alternatives:

1<<np.arange(17)
2**np.arange(17)
np.left_shift.accumulate(np.ones(17,int))
np.repeat((1,2),(1,16)).cumprod()
np.vander([2],17,True)[0]
np.ldexp(1,np.arange(17),dtype=float)

Silly alternatives:

from scipy.sparse import linalg,diags
linalg.spsolve(diags([(1,),(-2,)],(0,-1),(17,17)),np.r_[:17]==0    
np.packbits(np.identity(32,'?')[:17],1,'little').view('<i4').T[0] 
np.ravel_multi_index(np.identity(17,int)[::-1],np.full(17,2))
np.where(np.sum(np.ix_(*17*((0,1),))).reshape(-1)==1)[0]

edited Sep 18, 2019 at 9:49

answered Sep 18, 2019 at 8:17

Paul Panzer

53.3k3 gold badges59 silver badges103 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

crayxt · Accepted Answer · 2019-09-18 04:03:53Z

0

You need to assign the value back

import numpy as np


array = np.zeros(shape=17, dtype="int")

x = 1

for i in range(len(array)):
    array[i] = x
    print(x)
    x *= 2

>>> print(array)
[    1     2     4     8    16    32    64   128   256   512  1024  2048
  4096  8192 16384 32768 65536]

answered Sep 18, 2019 at 4:03

crayxt

2,4052 gold badges15 silver badges19 bronze badges

Comments

Dev Khadka · Accepted Answer · 2019-09-18 04:37:49Z

0

it will be more efficient using numpy vectorization like below.

import numpy as np
n=17
triangle = (np.tri(n,n,-1, dtype=np.int64)+1)
triangle.cumprod(axis=1)[:,-1]

Explanation

np.tri(n,n, dtype=np.int64) will create triangle matrix with values 1 at and below diagonal and 0 else where

np.tri(n,n, -1, dtype=np.int64) will shift the triangle matrix by one row such that first row is all zero

np.tri(n,n, -1, dtype=np.int64)+1 will change 0s to 1s and 1s to 2s

at last step use cumprod and take last column which is our answer as it will be products of 0,1,2 ... n 2's with remaining 1s

edited Sep 18, 2019 at 4:37

answered Sep 18, 2019 at 4:32

Dev Khadka

5,5415 gold badges23 silver badges36 bronze badges

Collectives™ on Stack Overflow

Convert for loop outcome to numpy array

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related