findall() or filter() functions with multiple conditions in julia 1.0.3
I am trying to convert following python codes to julia 1.0.3
from numpy import *
xl,xr,yl,yr = 0,1,0,1
xs,ys = linspace(xl,xr,N),linspace(yl,yr,N)
x,y = np.meshgrid(xs,ys)
data=column_stack((ravel(x),ravel(y)))
idx1 = where((data[:,0]==xl) | (data[:,0]==xr) | (data[:,1]==yl) | (data[:,1]==yr))
I could not convert last row of the above codes. I came across findall() and filter() functions but could not use them properly in my case
julia-lang
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
add a comment |
I am trying to convert following python codes to julia 1.0.3
from numpy import *
xl,xr,yl,yr = 0,1,0,1
xs,ys = linspace(xl,xr,N),linspace(yl,yr,N)
x,y = np.meshgrid(xs,ys)
data=column_stack((ravel(x),ravel(y)))
idx1 = where((data[:,0]==xl) | (data[:,0]==xr) | (data[:,1]==yl) | (data[:,1]==yr))
I could not convert last row of the above codes. I came across findall() and filter() functions but could not use them properly in my case
julia-lang
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
Can you explain what your code should do? Not everyone is familiar with numpy, and thiswhereline is not so easy to interpret.
– phg
Dec 28 '18 at 19:44
add a comment |
I am trying to convert following python codes to julia 1.0.3
from numpy import *
xl,xr,yl,yr = 0,1,0,1
xs,ys = linspace(xl,xr,N),linspace(yl,yr,N)
x,y = np.meshgrid(xs,ys)
data=column_stack((ravel(x),ravel(y)))
idx1 = where((data[:,0]==xl) | (data[:,0]==xr) | (data[:,1]==yl) | (data[:,1]==yr))
I could not convert last row of the above codes. I came across findall() and filter() functions but could not use them properly in my case
julia-lang
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
I am trying to convert following python codes to julia 1.0.3
from numpy import *
xl,xr,yl,yr = 0,1,0,1
xs,ys = linspace(xl,xr,N),linspace(yl,yr,N)
x,y = np.meshgrid(xs,ys)
data=column_stack((ravel(x),ravel(y)))
idx1 = where((data[:,0]==xl) | (data[:,0]==xr) | (data[:,1]==yl) | (data[:,1]==yr))
I could not convert last row of the above codes. I came across findall() and filter() functions but could not use them properly in my case
julia-lang
julia-lang
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
edited Dec 28 '18 at 19:36
user1772257
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
asked Dec 28 '18 at 17:53
user1772257user1772257
178212
178212
Can you explain what your code should do? Not everyone is familiar with numpy, and thiswhereline is not so easy to interpret.
– phg
Dec 28 '18 at 19:44
add a comment |
Can you explain what your code should do? Not everyone is familiar with numpy, and thiswhereline is not so easy to interpret.
– phg
Dec 28 '18 at 19:44
Can you explain what your code should do? Not everyone is familiar with numpy, and this
where line is not so easy to interpret.– phg
Dec 28 '18 at 19:44
Can you explain what your code should do? Not everyone is familiar with numpy, and this
where line is not so easy to interpret.– phg
Dec 28 '18 at 19:44
add a comment |
1 Answer
1
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oldest
votes
I understand you want data to hold all pairs you can create from elements of xs and ys and select the indices of elements of data that are on the border to idx1. If this is the case this is how I would implement it in Julia:
n=11
xl, xr, yl, yr = 0,1,0,1
xs, ys = range(xl, stop=xr, length=n), range(xl, stop=xr, length=n)
data = [(x,y) for y in ys for x in xs]
idx1 = findall(((x,y),) -> x in (xl,xr) || y in (yl, yr), data)
If you want data to be a matrix not a vector you could do:
data2 = reduce(vcat, [x y] for y in ys for x in xs)
idx12 = filter(i -> data2[i,1] in (xl,xr) || data2[i,2] in (yl, yr), axes(data2, 1))
but in this case for me it would be more natural in Julia to use a vector of tuples rather than a matrix.
You can also consider using Iterators.product function to generate data like this vec(collect(Iterators.product(xs, ys))).
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
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votes
I understand you want data to hold all pairs you can create from elements of xs and ys and select the indices of elements of data that are on the border to idx1. If this is the case this is how I would implement it in Julia:
n=11
xl, xr, yl, yr = 0,1,0,1
xs, ys = range(xl, stop=xr, length=n), range(xl, stop=xr, length=n)
data = [(x,y) for y in ys for x in xs]
idx1 = findall(((x,y),) -> x in (xl,xr) || y in (yl, yr), data)
If you want data to be a matrix not a vector you could do:
data2 = reduce(vcat, [x y] for y in ys for x in xs)
idx12 = filter(i -> data2[i,1] in (xl,xr) || data2[i,2] in (yl, yr), axes(data2, 1))
but in this case for me it would be more natural in Julia to use a vector of tuples rather than a matrix.
You can also consider using Iterators.product function to generate data like this vec(collect(Iterators.product(xs, ys))).
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
add a comment |
I understand you want data to hold all pairs you can create from elements of xs and ys and select the indices of elements of data that are on the border to idx1. If this is the case this is how I would implement it in Julia:
n=11
xl, xr, yl, yr = 0,1,0,1
xs, ys = range(xl, stop=xr, length=n), range(xl, stop=xr, length=n)
data = [(x,y) for y in ys for x in xs]
idx1 = findall(((x,y),) -> x in (xl,xr) || y in (yl, yr), data)
If you want data to be a matrix not a vector you could do:
data2 = reduce(vcat, [x y] for y in ys for x in xs)
idx12 = filter(i -> data2[i,1] in (xl,xr) || data2[i,2] in (yl, yr), axes(data2, 1))
but in this case for me it would be more natural in Julia to use a vector of tuples rather than a matrix.
You can also consider using Iterators.product function to generate data like this vec(collect(Iterators.product(xs, ys))).
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
add a comment |
I understand you want data to hold all pairs you can create from elements of xs and ys and select the indices of elements of data that are on the border to idx1. If this is the case this is how I would implement it in Julia:
n=11
xl, xr, yl, yr = 0,1,0,1
xs, ys = range(xl, stop=xr, length=n), range(xl, stop=xr, length=n)
data = [(x,y) for y in ys for x in xs]
idx1 = findall(((x,y),) -> x in (xl,xr) || y in (yl, yr), data)
If you want data to be a matrix not a vector you could do:
data2 = reduce(vcat, [x y] for y in ys for x in xs)
idx12 = filter(i -> data2[i,1] in (xl,xr) || data2[i,2] in (yl, yr), axes(data2, 1))
but in this case for me it would be more natural in Julia to use a vector of tuples rather than a matrix.
You can also consider using Iterators.product function to generate data like this vec(collect(Iterators.product(xs, ys))).
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
I understand you want data to hold all pairs you can create from elements of xs and ys and select the indices of elements of data that are on the border to idx1. If this is the case this is how I would implement it in Julia:
n=11
xl, xr, yl, yr = 0,1,0,1
xs, ys = range(xl, stop=xr, length=n), range(xl, stop=xr, length=n)
data = [(x,y) for y in ys for x in xs]
idx1 = findall(((x,y),) -> x in (xl,xr) || y in (yl, yr), data)
If you want data to be a matrix not a vector you could do:
data2 = reduce(vcat, [x y] for y in ys for x in xs)
idx12 = filter(i -> data2[i,1] in (xl,xr) || data2[i,2] in (yl, yr), axes(data2, 1))
but in this case for me it would be more natural in Julia to use a vector of tuples rather than a matrix.
You can also consider using Iterators.product function to generate data like this vec(collect(Iterators.product(xs, ys))).
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
edited Dec 28 '18 at 20:39
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
answered Dec 28 '18 at 20:24
Bogumił KamińskiBogumił Kamiński
12.3k11120
12.3k11120
add a comment |
add a comment |
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Can you explain what your code should do? Not everyone is familiar with numpy, and this
whereline is not so easy to interpret.– phg
Dec 28 '18 at 19:44