Replace multiple elements in numpy array with 1

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up vote
9
down vote

favorite
3












In a given numpy array X:



X = array([1,2,3,4,5,6,7,8,9,10])


I would like to replace indices (2, 3) and (7, 8) with a single element -1 respectively, like:



X = array([1,2,-1,5,6,7,-1,10])


In other words, I replaced values at indices (2, 3) and (7,8) of the original array with a singular value.



Question is: Is there a numpy-ish way (i.e. without for loops and usage of python lists) around it? Thanks.



Note: This is NOT equivalent of replacing a single element in-place with another. Its about replacing multiple values with a "singular" value. Thanks.






python arrays numpy







share|improve this question



















  • 2




    you don't have to be rude, especially when you are referring to a manual!
    – khan
    1 hour ago










  • Possible duplicate of Replace an element in a numpy array at specific index
    – awiebe
    1 hour ago






  • 1




    You should google then, if you have a different meaning in your mind.
    – khan
    1 hour ago






  • 1




    @awiebe Not sure what this means ... the question is looking for a numpy-ish way of replacing two entities in an array with a single entity (demo-ed by the examples too in which you can clearly see the size of array reduced by 2 elements).
    – khan
    1 hour ago






  • 1




    Ah, I see the question was resizing the array, not replacing the value. That makes more sense then how I read it.
    – awiebe
    1 hour ago














up vote
9
down vote

favorite
3












In a given numpy array X:



X = array([1,2,3,4,5,6,7,8,9,10])


I would like to replace indices (2, 3) and (7, 8) with a single element -1 respectively, like:



X = array([1,2,-1,5,6,7,-1,10])


In other words, I replaced values at indices (2, 3) and (7,8) of the original array with a singular value.



Question is: Is there a numpy-ish way (i.e. without for loops and usage of python lists) around it? Thanks.



Note: This is NOT equivalent of replacing a single element in-place with another. Its about replacing multiple values with a "singular" value. Thanks.






python arrays numpy







share|improve this question



















  • 2




    you don't have to be rude, especially when you are referring to a manual!
    – khan
    1 hour ago










  • Possible duplicate of Replace an element in a numpy array at specific index
    – awiebe
    1 hour ago






  • 1




    You should google then, if you have a different meaning in your mind.
    – khan
    1 hour ago






  • 1




    @awiebe Not sure what this means ... the question is looking for a numpy-ish way of replacing two entities in an array with a single entity (demo-ed by the examples too in which you can clearly see the size of array reduced by 2 elements).
    – khan
    1 hour ago






  • 1




    Ah, I see the question was resizing the array, not replacing the value. That makes more sense then how I read it.
    – awiebe
    1 hour ago












up vote
9
down vote

favorite
3









up vote
9
down vote

favorite
3






3





In a given numpy array X:



X = array([1,2,3,4,5,6,7,8,9,10])


I would like to replace indices (2, 3) and (7, 8) with a single element -1 respectively, like:



X = array([1,2,-1,5,6,7,-1,10])


In other words, I replaced values at indices (2, 3) and (7,8) of the original array with a singular value.



Question is: Is there a numpy-ish way (i.e. without for loops and usage of python lists) around it? Thanks.



Note: This is NOT equivalent of replacing a single element in-place with another. Its about replacing multiple values with a "singular" value. Thanks.






python arrays numpy







share|improve this question















In a given numpy array X:



X = array([1,2,3,4,5,6,7,8,9,10])


I would like to replace indices (2, 3) and (7, 8) with a single element -1 respectively, like:



X = array([1,2,-1,5,6,7,-1,10])


In other words, I replaced values at indices (2, 3) and (7,8) of the original array with a singular value.



Question is: Is there a numpy-ish way (i.e. without for loops and usage of python lists) around it? Thanks.



Note: This is NOT equivalent of replacing a single element in-place with another. Its about replacing multiple values with a "singular" value. Thanks.






python arrays numpy




python arrays numpy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 35 mins ago









user3483203

27.4k72350




27.4k72350










asked 2 hours ago









khan

1,66583049




1,66583049







  • 2




    you don't have to be rude, especially when you are referring to a manual!
    – khan
    1 hour ago










  • Possible duplicate of Replace an element in a numpy array at specific index
    – awiebe
    1 hour ago






  • 1




    You should google then, if you have a different meaning in your mind.
    – khan
    1 hour ago






  • 1




    @awiebe Not sure what this means ... the question is looking for a numpy-ish way of replacing two entities in an array with a single entity (demo-ed by the examples too in which you can clearly see the size of array reduced by 2 elements).
    – khan
    1 hour ago






  • 1




    Ah, I see the question was resizing the array, not replacing the value. That makes more sense then how I read it.
    – awiebe
    1 hour ago












  • 2




    you don't have to be rude, especially when you are referring to a manual!
    – khan
    1 hour ago










  • Possible duplicate of Replace an element in a numpy array at specific index
    – awiebe
    1 hour ago






  • 1




    You should google then, if you have a different meaning in your mind.
    – khan
    1 hour ago






  • 1




    @awiebe Not sure what this means ... the question is looking for a numpy-ish way of replacing two entities in an array with a single entity (demo-ed by the examples too in which you can clearly see the size of array reduced by 2 elements).
    – khan
    1 hour ago






  • 1




    Ah, I see the question was resizing the array, not replacing the value. That makes more sense then how I read it.
    – awiebe
    1 hour ago







2




2




you don't have to be rude, especially when you are referring to a manual!
– khan
1 hour ago




you don't have to be rude, especially when you are referring to a manual!
– khan
1 hour ago












Possible duplicate of Replace an element in a numpy array at specific index
– awiebe
1 hour ago




Possible duplicate of Replace an element in a numpy array at specific index
– awiebe
1 hour ago




1




1




You should google then, if you have a different meaning in your mind.
– khan
1 hour ago




You should google then, if you have a different meaning in your mind.
– khan
1 hour ago




1




1




@awiebe Not sure what this means ... the question is looking for a numpy-ish way of replacing two entities in an array with a single entity (demo-ed by the examples too in which you can clearly see the size of array reduced by 2 elements).
– khan
1 hour ago




@awiebe Not sure what this means ... the question is looking for a numpy-ish way of replacing two entities in an array with a single entity (demo-ed by the examples too in which you can clearly see the size of array reduced by 2 elements).
– khan
1 hour ago




1




1




Ah, I see the question was resizing the array, not replacing the value. That makes more sense then how I read it.
– awiebe
1 hour ago




Ah, I see the question was resizing the array, not replacing the value. That makes more sense then how I read it.
– awiebe
1 hour ago












3 Answers
3






active

oldest

votes

















up vote
0
down vote



accepted










Try np.put:



np.put(X, [2,3,7,8], [-1,0]) # `0` can be changed to anything that's not in the array
print(X[X!=0]) # whatever You put as an number in `put`


So basically use put to do the values for the indexes, then drop the zero-values.



Or as @khan says, can do something that's out of range:



np.put(X, [2,3,7,8], [-1,np.max(X)+1])
print(X[X!=X.max()])


All Output:



[ 1 2 -1 5 6 7 -1 10]





share|improve this answer


















  • 2




    what if the list contains 0 initially?
    – Ashutosh Chapagain
    1 hour ago






  • 2




    @AshutoshChapagain Oh yeah smart..., i'll edit
    – U9-Forward
    1 hour ago






  • 1




    This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
    – khan
    1 hour ago






  • 1




    Ohh you already edited. Great answer. :-)
    – khan
    1 hour ago






  • 1




    A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
    – khan
    1 hour ago

















up vote
3
down vote













I'm not sure if this can be done in one step, but here's a way using np.delete:



import numpy as np
from operator import itemgetter

X = np.array(range(1,11))
to_replace = [[2,3], [7,8]]


X[list(map(itemgetter(0), to_replace))] = -1
X = np.delete(X, list(map(lambda x: x[1:], to_replace)))
print(X)
#[ 1 2 -1 5 6 7 -1 10]


First we replace the first element of each pair with -1. Then we delete the remaining elements.






share|improve this answer






















  • Amm.. this is not fully a numpy-ish solution tho...
    – U9-Forward
    1 hour ago

















up vote
3
down vote













A solution using numpy.delete, similar to @pault, but more efficient as it uses pure numpy indexing. However, because of this efficient indexing, it means that you cannot pass jagged arrays as indices



Setup



a = np.array([1,2,3,4,5,6,7,8,9,10])
idx = np.stack([[2, 3], [7, 8]])


a[idx] = -1
np.delete(a, idx[:, 1:])




array([ 1, 2, -1, 5, 6, 7, -1, 10])





share|improve this answer






















  • Definitely seems like a cleaner and more efficient way.
    – BernardL
    53 mins ago










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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Try np.put:



np.put(X, [2,3,7,8], [-1,0]) # `0` can be changed to anything that's not in the array
print(X[X!=0]) # whatever You put as an number in `put`


So basically use put to do the values for the indexes, then drop the zero-values.



Or as @khan says, can do something that's out of range:



np.put(X, [2,3,7,8], [-1,np.max(X)+1])
print(X[X!=X.max()])


All Output:



[ 1 2 -1 5 6 7 -1 10]





share|improve this answer


















  • 2




    what if the list contains 0 initially?
    – Ashutosh Chapagain
    1 hour ago






  • 2




    @AshutoshChapagain Oh yeah smart..., i'll edit
    – U9-Forward
    1 hour ago






  • 1




    This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
    – khan
    1 hour ago






  • 1




    Ohh you already edited. Great answer. :-)
    – khan
    1 hour ago






  • 1




    A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
    – khan
    1 hour ago














up vote
0
down vote



accepted










Try np.put:



np.put(X, [2,3,7,8], [-1,0]) # `0` can be changed to anything that's not in the array
print(X[X!=0]) # whatever You put as an number in `put`


So basically use put to do the values for the indexes, then drop the zero-values.



Or as @khan says, can do something that's out of range:



np.put(X, [2,3,7,8], [-1,np.max(X)+1])
print(X[X!=X.max()])


All Output:



[ 1 2 -1 5 6 7 -1 10]





share|improve this answer


















  • 2




    what if the list contains 0 initially?
    – Ashutosh Chapagain
    1 hour ago






  • 2




    @AshutoshChapagain Oh yeah smart..., i'll edit
    – U9-Forward
    1 hour ago






  • 1




    This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
    – khan
    1 hour ago






  • 1




    Ohh you already edited. Great answer. :-)
    – khan
    1 hour ago






  • 1




    A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
    – khan
    1 hour ago












up vote
0
down vote



accepted







up vote
0
down vote



accepted






Try np.put:



np.put(X, [2,3,7,8], [-1,0]) # `0` can be changed to anything that's not in the array
print(X[X!=0]) # whatever You put as an number in `put`


So basically use put to do the values for the indexes, then drop the zero-values.



Or as @khan says, can do something that's out of range:



np.put(X, [2,3,7,8], [-1,np.max(X)+1])
print(X[X!=X.max()])


All Output:



[ 1 2 -1 5 6 7 -1 10]





share|improve this answer














Try np.put:



np.put(X, [2,3,7,8], [-1,0]) # `0` can be changed to anything that's not in the array
print(X[X!=0]) # whatever You put as an number in `put`


So basically use put to do the values for the indexes, then drop the zero-values.



Or as @khan says, can do something that's out of range:



np.put(X, [2,3,7,8], [-1,np.max(X)+1])
print(X[X!=X.max()])


All Output:



[ 1 2 -1 5 6 7 -1 10]






share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 1 hour ago









U9-Forward

7,8342731




7,8342731







  • 2




    what if the list contains 0 initially?
    – Ashutosh Chapagain
    1 hour ago






  • 2




    @AshutoshChapagain Oh yeah smart..., i'll edit
    – U9-Forward
    1 hour ago






  • 1




    This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
    – khan
    1 hour ago






  • 1




    Ohh you already edited. Great answer. :-)
    – khan
    1 hour ago






  • 1




    A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
    – khan
    1 hour ago












  • 2




    what if the list contains 0 initially?
    – Ashutosh Chapagain
    1 hour ago






  • 2




    @AshutoshChapagain Oh yeah smart..., i'll edit
    – U9-Forward
    1 hour ago






  • 1




    This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
    – khan
    1 hour ago






  • 1




    Ohh you already edited. Great answer. :-)
    – khan
    1 hour ago






  • 1




    A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
    – khan
    1 hour ago







2




2




what if the list contains 0 initially?
– Ashutosh Chapagain
1 hour ago




what if the list contains 0 initially?
– Ashutosh Chapagain
1 hour ago




2




2




@AshutoshChapagain Oh yeah smart..., i'll edit
– U9-Forward
1 hour ago




@AshutoshChapagain Oh yeah smart..., i'll edit
– U9-Forward
1 hour ago




1




1




This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
– khan
1 hour ago




This zero can actually be anything out of the range of the array..that will help it. So, X=numpy.arange(1,11); numpy.put(X, [2,3,7,8], [-1,numpy.max(X)+1]); X[X!=X.max()]..
– khan
1 hour ago




1




1




Ohh you already edited. Great answer. :-)
– khan
1 hour ago




Ohh you already edited. Great answer. :-)
– khan
1 hour ago




1




1




A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
– khan
1 hour ago




A food for thought: What if we like to replace indices (2,3,7,8) with -1 and indices (9,10) with -2..
– khan
1 hour ago












up vote
3
down vote













I'm not sure if this can be done in one step, but here's a way using np.delete:



import numpy as np
from operator import itemgetter

X = np.array(range(1,11))
to_replace = [[2,3], [7,8]]


X[list(map(itemgetter(0), to_replace))] = -1
X = np.delete(X, list(map(lambda x: x[1:], to_replace)))
print(X)
#[ 1 2 -1 5 6 7 -1 10]


First we replace the first element of each pair with -1. Then we delete the remaining elements.






share|improve this answer






















  • Amm.. this is not fully a numpy-ish solution tho...
    – U9-Forward
    1 hour ago














up vote
3
down vote













I'm not sure if this can be done in one step, but here's a way using np.delete:



import numpy as np
from operator import itemgetter

X = np.array(range(1,11))
to_replace = [[2,3], [7,8]]


X[list(map(itemgetter(0), to_replace))] = -1
X = np.delete(X, list(map(lambda x: x[1:], to_replace)))
print(X)
#[ 1 2 -1 5 6 7 -1 10]


First we replace the first element of each pair with -1. Then we delete the remaining elements.






share|improve this answer






















  • Amm.. this is not fully a numpy-ish solution tho...
    – U9-Forward
    1 hour ago












up vote
3
down vote










up vote
3
down vote









I'm not sure if this can be done in one step, but here's a way using np.delete:



import numpy as np
from operator import itemgetter

X = np.array(range(1,11))
to_replace = [[2,3], [7,8]]


X[list(map(itemgetter(0), to_replace))] = -1
X = np.delete(X, list(map(lambda x: x[1:], to_replace)))
print(X)
#[ 1 2 -1 5 6 7 -1 10]


First we replace the first element of each pair with -1. Then we delete the remaining elements.






share|improve this answer














I'm not sure if this can be done in one step, but here's a way using np.delete:



import numpy as np
from operator import itemgetter

X = np.array(range(1,11))
to_replace = [[2,3], [7,8]]


X[list(map(itemgetter(0), to_replace))] = -1
X = np.delete(X, list(map(lambda x: x[1:], to_replace)))
print(X)
#[ 1 2 -1 5 6 7 -1 10]


First we replace the first element of each pair with -1. Then we delete the remaining elements.







share|improve this answer














share|improve this answer



share|improve this answer








edited 59 mins ago

























answered 1 hour ago









pault

12.6k31541




12.6k31541











  • Amm.. this is not fully a numpy-ish solution tho...
    – U9-Forward
    1 hour ago
















  • Amm.. this is not fully a numpy-ish solution tho...
    – U9-Forward
    1 hour ago















Amm.. this is not fully a numpy-ish solution tho...
– U9-Forward
1 hour ago




Amm.. this is not fully a numpy-ish solution tho...
– U9-Forward
1 hour ago










up vote
3
down vote













A solution using numpy.delete, similar to @pault, but more efficient as it uses pure numpy indexing. However, because of this efficient indexing, it means that you cannot pass jagged arrays as indices



Setup



a = np.array([1,2,3,4,5,6,7,8,9,10])
idx = np.stack([[2, 3], [7, 8]])


a[idx] = -1
np.delete(a, idx[:, 1:])




array([ 1, 2, -1, 5, 6, 7, -1, 10])





share|improve this answer






















  • Definitely seems like a cleaner and more efficient way.
    – BernardL
    53 mins ago














up vote
3
down vote













A solution using numpy.delete, similar to @pault, but more efficient as it uses pure numpy indexing. However, because of this efficient indexing, it means that you cannot pass jagged arrays as indices



Setup



a = np.array([1,2,3,4,5,6,7,8,9,10])
idx = np.stack([[2, 3], [7, 8]])


a[idx] = -1
np.delete(a, idx[:, 1:])




array([ 1, 2, -1, 5, 6, 7, -1, 10])





share|improve this answer






















  • Definitely seems like a cleaner and more efficient way.
    – BernardL
    53 mins ago












up vote
3
down vote










up vote
3
down vote









A solution using numpy.delete, similar to @pault, but more efficient as it uses pure numpy indexing. However, because of this efficient indexing, it means that you cannot pass jagged arrays as indices



Setup



a = np.array([1,2,3,4,5,6,7,8,9,10])
idx = np.stack([[2, 3], [7, 8]])


a[idx] = -1
np.delete(a, idx[:, 1:])




array([ 1, 2, -1, 5, 6, 7, -1, 10])





share|improve this answer














A solution using numpy.delete, similar to @pault, but more efficient as it uses pure numpy indexing. However, because of this efficient indexing, it means that you cannot pass jagged arrays as indices



Setup



a = np.array([1,2,3,4,5,6,7,8,9,10])
idx = np.stack([[2, 3], [7, 8]])


a[idx] = -1
np.delete(a, idx[:, 1:])




array([ 1, 2, -1, 5, 6, 7, -1, 10])






share|improve this answer














share|improve this answer



share|improve this answer








edited 53 mins ago

























answered 55 mins ago









user3483203

27.4k72350




27.4k72350











  • Definitely seems like a cleaner and more efficient way.
    – BernardL
    53 mins ago
















  • Definitely seems like a cleaner and more efficient way.
    – BernardL
    53 mins ago















Definitely seems like a cleaner and more efficient way.
– BernardL
53 mins ago




Definitely seems like a cleaner and more efficient way.
– BernardL
53 mins ago

















 

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Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
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Confectionery

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Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
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