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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Installing NextGIS Connect into QGIS 3?

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Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
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One-line joke

A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
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