How to loop trough elements of an array wihout a for loop in python?











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-1
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sindrometestearray[len(sindrometestearray)]
gives the [idxOfElementToChange,idxOfBitToChange]



I have these arrays:



sindrometestearray = [[0,0],[2,1]] 
c1divided = [[0,0,0,0],[0,0,0,0],[0,0,0,0]]


example: sindrometestearray[0] = [0,0], in c1divided, the first bit of the first element should be the opposite.
like : [1,0,0,1]



for n in sindrometestearray: 
c1divided[n[0]][n[1]] = not(c1divided[n[0]][n[1]])


Output:



[[1,0,0,0],[0,0,0,0],[0,1,0,0]]


How could I do it without for loops?










share|improve this question
























  • Are you (supposed to be) using NumPy?
    – user2357112
    Nov 21 at 23:42










  • Yes. Im importing numpy as np
    – Yunrin 127
    Nov 21 at 23:43










  • Welcome to SO, it’s great that you’ve included your code and intended result, could you please explain in more detail what exactly it’s supposed to do to get that?
    – Jaba
    Nov 21 at 23:46










  • Shouldn't your output be [[True, 1, 0, 1], [0, 1, 1, 0], [0, True, 0, 0]] instead?
    – blhsing
    Nov 21 at 23:50










  • @blhsing yes the True should appear as 1, but yes.
    – Yunrin 127
    Nov 21 at 23:53















up vote
-1
down vote

favorite












sindrometestearray[len(sindrometestearray)]
gives the [idxOfElementToChange,idxOfBitToChange]



I have these arrays:



sindrometestearray = [[0,0],[2,1]] 
c1divided = [[0,0,0,0],[0,0,0,0],[0,0,0,0]]


example: sindrometestearray[0] = [0,0], in c1divided, the first bit of the first element should be the opposite.
like : [1,0,0,1]



for n in sindrometestearray: 
c1divided[n[0]][n[1]] = not(c1divided[n[0]][n[1]])


Output:



[[1,0,0,0],[0,0,0,0],[0,1,0,0]]


How could I do it without for loops?










share|improve this question
























  • Are you (supposed to be) using NumPy?
    – user2357112
    Nov 21 at 23:42










  • Yes. Im importing numpy as np
    – Yunrin 127
    Nov 21 at 23:43










  • Welcome to SO, it’s great that you’ve included your code and intended result, could you please explain in more detail what exactly it’s supposed to do to get that?
    – Jaba
    Nov 21 at 23:46










  • Shouldn't your output be [[True, 1, 0, 1], [0, 1, 1, 0], [0, True, 0, 0]] instead?
    – blhsing
    Nov 21 at 23:50










  • @blhsing yes the True should appear as 1, but yes.
    – Yunrin 127
    Nov 21 at 23:53













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











sindrometestearray[len(sindrometestearray)]
gives the [idxOfElementToChange,idxOfBitToChange]



I have these arrays:



sindrometestearray = [[0,0],[2,1]] 
c1divided = [[0,0,0,0],[0,0,0,0],[0,0,0,0]]


example: sindrometestearray[0] = [0,0], in c1divided, the first bit of the first element should be the opposite.
like : [1,0,0,1]



for n in sindrometestearray: 
c1divided[n[0]][n[1]] = not(c1divided[n[0]][n[1]])


Output:



[[1,0,0,0],[0,0,0,0],[0,1,0,0]]


How could I do it without for loops?










share|improve this question















sindrometestearray[len(sindrometestearray)]
gives the [idxOfElementToChange,idxOfBitToChange]



I have these arrays:



sindrometestearray = [[0,0],[2,1]] 
c1divided = [[0,0,0,0],[0,0,0,0],[0,0,0,0]]


example: sindrometestearray[0] = [0,0], in c1divided, the first bit of the first element should be the opposite.
like : [1,0,0,1]



for n in sindrometestearray: 
c1divided[n[0]][n[1]] = not(c1divided[n[0]][n[1]])


Output:



[[1,0,0,0],[0,0,0,0],[0,1,0,0]]


How could I do it without for loops?







python






share|improve this question















share|improve this question













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share|improve this question








edited Nov 22 at 9:50

























asked Nov 21 at 23:41









Yunrin 127

41




41












  • Are you (supposed to be) using NumPy?
    – user2357112
    Nov 21 at 23:42










  • Yes. Im importing numpy as np
    – Yunrin 127
    Nov 21 at 23:43










  • Welcome to SO, it’s great that you’ve included your code and intended result, could you please explain in more detail what exactly it’s supposed to do to get that?
    – Jaba
    Nov 21 at 23:46










  • Shouldn't your output be [[True, 1, 0, 1], [0, 1, 1, 0], [0, True, 0, 0]] instead?
    – blhsing
    Nov 21 at 23:50










  • @blhsing yes the True should appear as 1, but yes.
    – Yunrin 127
    Nov 21 at 23:53


















  • Are you (supposed to be) using NumPy?
    – user2357112
    Nov 21 at 23:42










  • Yes. Im importing numpy as np
    – Yunrin 127
    Nov 21 at 23:43










  • Welcome to SO, it’s great that you’ve included your code and intended result, could you please explain in more detail what exactly it’s supposed to do to get that?
    – Jaba
    Nov 21 at 23:46










  • Shouldn't your output be [[True, 1, 0, 1], [0, 1, 1, 0], [0, True, 0, 0]] instead?
    – blhsing
    Nov 21 at 23:50










  • @blhsing yes the True should appear as 1, but yes.
    – Yunrin 127
    Nov 21 at 23:53
















Are you (supposed to be) using NumPy?
– user2357112
Nov 21 at 23:42




Are you (supposed to be) using NumPy?
– user2357112
Nov 21 at 23:42












Yes. Im importing numpy as np
– Yunrin 127
Nov 21 at 23:43




Yes. Im importing numpy as np
– Yunrin 127
Nov 21 at 23:43












Welcome to SO, it’s great that you’ve included your code and intended result, could you please explain in more detail what exactly it’s supposed to do to get that?
– Jaba
Nov 21 at 23:46




Welcome to SO, it’s great that you’ve included your code and intended result, could you please explain in more detail what exactly it’s supposed to do to get that?
– Jaba
Nov 21 at 23:46












Shouldn't your output be [[True, 1, 0, 1], [0, 1, 1, 0], [0, True, 0, 0]] instead?
– blhsing
Nov 21 at 23:50




Shouldn't your output be [[True, 1, 0, 1], [0, 1, 1, 0], [0, True, 0, 0]] instead?
– blhsing
Nov 21 at 23:50












@blhsing yes the True should appear as 1, but yes.
– Yunrin 127
Nov 21 at 23:53




@blhsing yes the True should appear as 1, but yes.
– Yunrin 127
Nov 21 at 23:53

















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