changing array value one at a time
I am trying to flip a binary array one at a time.
import numpy as np
k = np.array([0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1])
for example my out put should be like this;
[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 1st output
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 2nd output
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 3rd output
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 4th output
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 5th output
In the first output, I want to flip only the first element of the array (other elements do not change), in the second output, the second element should change (the 1st and the remaining elements should not change).. etc.
Could anyone tell me how can flip one at a time? Thank you
python arrays python-3.x numpy
asked Nov 26 '18 at 20:02
Mass17Mass17
838
add a comment |
I am trying to flip a binary array one at a time.
import numpy as np
k = np.array([0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1])
for example my out put should be like this;
[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 1st output
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 2nd output
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 3rd output
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 4th output
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 5th output
In the first output, I want to flip only the first element of the array (other elements do not change), in the second output, the second element should change (the 1st and the remaining elements should not change).. etc.
Could anyone tell me how can flip one at a time? Thank you
python arrays python-3.x numpy
asked Nov 26 '18 at 20:02
Mass17Mass17
838
1
What have you tried so far?
– Ian Quah
Nov 26 '18 at 20:06
add a comment |
I am trying to flip a binary array one at a time.
import numpy as np
k = np.array([0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1])
for example my out put should be like this;
[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 1st output
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 2nd output
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 3rd output
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 4th output
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 5th output
In the first output, I want to flip only the first element of the array (other elements do not change), in the second output, the second element should change (the 1st and the remaining elements should not change).. etc.
Could anyone tell me how can flip one at a time? Thank you
python arrays python-3.x numpy
asked Nov 26 '18 at 20:02
Mass17Mass17
838
I am trying to flip a binary array one at a time.
import numpy as np
k = np.array([0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1])
for example my out put should be like this;
[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 1st output
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 2nd output
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 3rd output
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 4th output
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1] # 5th output
In the first output, I want to flip only the first element of the array (other elements do not change), in the second output, the second element should change (the 1st and the remaining elements should not change).. etc.
Could anyone tell me how can flip one at a time? Thank you
python arrays python-3.x numpy
python arrays python-3.x numpy
asked Nov 26 '18 at 20:02
Mass17Mass17
838
asked Nov 26 '18 at 20:02
Mass17Mass17
838
asked Nov 26 '18 at 20:02
Mass17Mass17
838
asked Nov 26 '18 at 20:02
Mass17Mass17
838
asked Nov 26 '18 at 20:02
Mass17Mass17
838
838
1
What have you tried so far?
– Ian Quah
Nov 26 '18 at 20:06
add a comment |
1
What have you tried so far?
– Ian Quah
Nov 26 '18 at 20:06
1
1
What have you tried so far?
– Ian Quah
Nov 26 '18 at 20:06
What have you tried so far?
– Ian Quah
Nov 26 '18 at 20:06
add a comment |
2 Answers
2
active
oldest
votes
What you're describing is flipping the diagonal of a tiled version of your array. By stacking your array, you can operate on the entire array at once, using vectorized operations, rather than operating on each row individually.
Setup
arr = np.tile(k, 5).reshape(-1, k.shape[0])
array([[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
Using numpy.diag_indices:
x, y = np.diag_indices(arr.shape[0])
arr[x, y] = 1 - arr[x, y]
array([[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
1
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
add a comment |
You can use a generator to save memory and even time on big arrays :
k=np.array([0, 0, 1])
def flip_one(k):
k[0]=1-k[0]
yield k
for i in range(len(k)):
k[i:i+2]=1-k[i:i+2]
yield k
for f in flip_one(k) :
print (f) # or other useful things!
#[1 0 1]
#[0 1 1]
#[0 0 0]
#[0 0 1]
k is reset at the end of the loop.
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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oldest
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active
oldest
votes
What you're describing is flipping the diagonal of a tiled version of your array. By stacking your array, you can operate on the entire array at once, using vectorized operations, rather than operating on each row individually.
Setup
arr = np.tile(k, 5).reshape(-1, k.shape[0])
array([[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
Using numpy.diag_indices:
x, y = np.diag_indices(arr.shape[0])
arr[x, y] = 1 - arr[x, y]
array([[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
1
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
add a comment |
What you're describing is flipping the diagonal of a tiled version of your array. By stacking your array, you can operate on the entire array at once, using vectorized operations, rather than operating on each row individually.
Setup
arr = np.tile(k, 5).reshape(-1, k.shape[0])
array([[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
Using numpy.diag_indices:
x, y = np.diag_indices(arr.shape[0])
arr[x, y] = 1 - arr[x, y]
array([[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
1
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
add a comment |
What you're describing is flipping the diagonal of a tiled version of your array. By stacking your array, you can operate on the entire array at once, using vectorized operations, rather than operating on each row individually.
Setup
arr = np.tile(k, 5).reshape(-1, k.shape[0])
array([[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
Using numpy.diag_indices:
x, y = np.diag_indices(arr.shape[0])
arr[x, y] = 1 - arr[x, y]
array([[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
What you're describing is flipping the diagonal of a tiled version of your array. By stacking your array, you can operate on the entire array at once, using vectorized operations, rather than operating on each row individually.
Setup
arr = np.tile(k, 5).reshape(-1, k.shape[0])
array([[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
Using numpy.diag_indices:
x, y = np.diag_indices(arr.shape[0])
arr[x, y] = 1 - arr[x, y]
array([[1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1]])
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
answered Nov 26 '18 at 20:11
user3483203user3483203
31.2k82655
31.2k82655
1
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
add a comment |
1
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
1
1
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
Thanks @user3483203. Very helpful technique!!
– Mass17
Nov 26 '18 at 20:19
add a comment |
You can use a generator to save memory and even time on big arrays :
k=np.array([0, 0, 1])
def flip_one(k):
k[0]=1-k[0]
yield k
for i in range(len(k)):
k[i:i+2]=1-k[i:i+2]
yield k
for f in flip_one(k) :
print (f) # or other useful things!
#[1 0 1]
#[0 1 1]
#[0 0 0]
#[0 0 1]
k is reset at the end of the loop.
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
add a comment |
You can use a generator to save memory and even time on big arrays :
k=np.array([0, 0, 1])
def flip_one(k):
k[0]=1-k[0]
yield k
for i in range(len(k)):
k[i:i+2]=1-k[i:i+2]
yield k
for f in flip_one(k) :
print (f) # or other useful things!
#[1 0 1]
#[0 1 1]
#[0 0 0]
#[0 0 1]
k is reset at the end of the loop.
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
add a comment |
You can use a generator to save memory and even time on big arrays :
k=np.array([0, 0, 1])
def flip_one(k):
k[0]=1-k[0]
yield k
for i in range(len(k)):
k[i:i+2]=1-k[i:i+2]
yield k
for f in flip_one(k) :
print (f) # or other useful things!
#[1 0 1]
#[0 1 1]
#[0 0 0]
#[0 0 1]
k is reset at the end of the loop.
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
You can use a generator to save memory and even time on big arrays :
k=np.array([0, 0, 1])
def flip_one(k):
k[0]=1-k[0]
yield k
for i in range(len(k)):
k[i:i+2]=1-k[i:i+2]
yield k
for f in flip_one(k) :
print (f) # or other useful things!
#[1 0 1]
#[0 1 1]
#[0 0 0]
#[0 0 1]
k is reset at the end of the loop.
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
edited Nov 26 '18 at 20:41
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
answered Nov 26 '18 at 20:35
B. M.B. M.
13.4k12034
13.4k12034
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
add a comment |
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
Thanks @B. M.. very helpful!!
– Mass17
Nov 26 '18 at 21:03
add a comment |
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What have you tried so far?
– Ian Quah
Nov 26 '18 at 20:06