How to replace Inf values with NaN in array in Julia 1.0?
I have seen online in a few places the solution
a = [1 2 3; 4 5 Inf]
a[isinf(a)] = NaN
But this gives me an error on Julia 1.0.1:
ERROR: MethodError: no method matching isinf(::Array{Float64,2})
Closest candidates are:
isinf(::BigFloat) at mpfr.jl:851
isinf(::Missing) at missing.jl:79
isinf(::ForwardDiff.Dual) at <path on my local machine>
What gives?
julia
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
add a comment |
I have seen online in a few places the solution
a = [1 2 3; 4 5 Inf]
a[isinf(a)] = NaN
But this gives me an error on Julia 1.0.1:
ERROR: MethodError: no method matching isinf(::Array{Float64,2})
Closest candidates are:
isinf(::BigFloat) at mpfr.jl:851
isinf(::Missing) at missing.jl:79
isinf(::ForwardDiff.Dual) at <path on my local machine>
What gives?
julia
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
add a comment |
I have seen online in a few places the solution
a = [1 2 3; 4 5 Inf]
a[isinf(a)] = NaN
But this gives me an error on Julia 1.0.1:
ERROR: MethodError: no method matching isinf(::Array{Float64,2})
Closest candidates are:
isinf(::BigFloat) at mpfr.jl:851
isinf(::Missing) at missing.jl:79
isinf(::ForwardDiff.Dual) at <path on my local machine>
What gives?
julia
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
I have seen online in a few places the solution
a = [1 2 3; 4 5 Inf]
a[isinf(a)] = NaN
But this gives me an error on Julia 1.0.1:
ERROR: MethodError: no method matching isinf(::Array{Float64,2})
Closest candidates are:
isinf(::BigFloat) at mpfr.jl:851
isinf(::Missing) at missing.jl:79
isinf(::ForwardDiff.Dual) at <path on my local machine>
What gives?
julia
julia
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
asked Nov 24 '18 at 23:49
nfernandnfernand
1201110
1201110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
As an additional comment. A standard function to perform this action is replace!. You can use it like this:
julia> a = [1 2 3; 4 5 Inf]
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 Inf
julia> replace!(a, Inf=>NaN)
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 NaN
It will perform better than broadcasting for large arrays.
If you really need speed you can write a simple function like this:
function inf2nan(x)
for i in eachindex(x)
@inbounds x[i] = ifelse(isinf(x[i]), NaN, x[i])
end
end
Now let us simply compare the performance of the three options:
julia> function bench()
x = fill(Inf, 10^8)
@time x[isinf.(x)] .= NaN
x = fill(Inf, 10^8)
@time replace!(x, Inf=>NaN)
x = fill(Inf, 10^8)
@time inf2nan(x)
end
bench (generic function with 1 method)
julia> bench()
0.980434 seconds (9 allocations: 774.865 MiB, 0.16% gc time)
0.183578 seconds
0.109929 seconds
julia> bench()
0.971408 seconds (9 allocations: 774.865 MiB, 0.03% gc time)
0.184163 seconds
0.102161 seconds
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
This is a very useful answer. I typed my answer without thinking about the fact thatisinf.(x)would allocate.
– Colin T Bowers
Nov 25 '18 at 8:40
2
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster thanreplace!in this case because we can avoid branching in the loop usingifelseand since the operation we perform is simple and@inboundsthe compiler can significantly optimize its execution time.
– Bogumił Kamiński
Nov 25 '18 at 8:50
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know whyisinfdoesn't work on arrays anymore.
– Colin T Bowers
Nov 25 '18 at 11:02
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
add a comment |
EDIT: For the most performant approaches to this problem see the excellent answer of @BogumilKaminski. This answer addresses the more general question of why isinf and related functions do not work on arrays anymore.
You are running into the more general issue that lots of functions that worked on arrays pre-v1.0 no longer work on arrays in v1.0 because you are supposed to be using broadcasting. The correct solution for v1.0 is:
a[isinf.(a)] .= NaN
I'm actually broadcasting in two places here. Firstly, we broadcast isinf over the array a, but we are also broadcasting the scalar NaN on the RHS to all indexed locations in the array on the LHS via .=. In general, the dot broadcasting notation is incredibly flexible and performant, and one of my favorite features of the latest iteration of Julia.
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
add a comment |
You are passing your entire array to isinf, it doesn't work on arrays, it works on numbers. Try this:
[isinf(i) ? NaN : i for i in a]
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
2
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
As an additional comment. A standard function to perform this action is replace!. You can use it like this:
julia> a = [1 2 3; 4 5 Inf]
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 Inf
julia> replace!(a, Inf=>NaN)
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 NaN
It will perform better than broadcasting for large arrays.
If you really need speed you can write a simple function like this:
function inf2nan(x)
for i in eachindex(x)
@inbounds x[i] = ifelse(isinf(x[i]), NaN, x[i])
end
end
Now let us simply compare the performance of the three options:
julia> function bench()
x = fill(Inf, 10^8)
@time x[isinf.(x)] .= NaN
x = fill(Inf, 10^8)
@time replace!(x, Inf=>NaN)
x = fill(Inf, 10^8)
@time inf2nan(x)
end
bench (generic function with 1 method)
julia> bench()
0.980434 seconds (9 allocations: 774.865 MiB, 0.16% gc time)
0.183578 seconds
0.109929 seconds
julia> bench()
0.971408 seconds (9 allocations: 774.865 MiB, 0.03% gc time)
0.184163 seconds
0.102161 seconds
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
This is a very useful answer. I typed my answer without thinking about the fact thatisinf.(x)would allocate.
– Colin T Bowers
Nov 25 '18 at 8:40
2
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster thanreplace!in this case because we can avoid branching in the loop usingifelseand since the operation we perform is simple and@inboundsthe compiler can significantly optimize its execution time.
– Bogumił Kamiński
Nov 25 '18 at 8:50
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know whyisinfdoesn't work on arrays anymore.
– Colin T Bowers
Nov 25 '18 at 11:02
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
add a comment |
As an additional comment. A standard function to perform this action is replace!. You can use it like this:
julia> a = [1 2 3; 4 5 Inf]
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 Inf
julia> replace!(a, Inf=>NaN)
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 NaN
It will perform better than broadcasting for large arrays.
If you really need speed you can write a simple function like this:
function inf2nan(x)
for i in eachindex(x)
@inbounds x[i] = ifelse(isinf(x[i]), NaN, x[i])
end
end
Now let us simply compare the performance of the three options:
julia> function bench()
x = fill(Inf, 10^8)
@time x[isinf.(x)] .= NaN
x = fill(Inf, 10^8)
@time replace!(x, Inf=>NaN)
x = fill(Inf, 10^8)
@time inf2nan(x)
end
bench (generic function with 1 method)
julia> bench()
0.980434 seconds (9 allocations: 774.865 MiB, 0.16% gc time)
0.183578 seconds
0.109929 seconds
julia> bench()
0.971408 seconds (9 allocations: 774.865 MiB, 0.03% gc time)
0.184163 seconds
0.102161 seconds
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
This is a very useful answer. I typed my answer without thinking about the fact thatisinf.(x)would allocate.
– Colin T Bowers
Nov 25 '18 at 8:40
2
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster thanreplace!in this case because we can avoid branching in the loop usingifelseand since the operation we perform is simple and@inboundsthe compiler can significantly optimize its execution time.
– Bogumił Kamiński
Nov 25 '18 at 8:50
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know whyisinfdoesn't work on arrays anymore.
– Colin T Bowers
Nov 25 '18 at 11:02
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
add a comment |
As an additional comment. A standard function to perform this action is replace!. You can use it like this:
julia> a = [1 2 3; 4 5 Inf]
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 Inf
julia> replace!(a, Inf=>NaN)
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 NaN
It will perform better than broadcasting for large arrays.
If you really need speed you can write a simple function like this:
function inf2nan(x)
for i in eachindex(x)
@inbounds x[i] = ifelse(isinf(x[i]), NaN, x[i])
end
end
Now let us simply compare the performance of the three options:
julia> function bench()
x = fill(Inf, 10^8)
@time x[isinf.(x)] .= NaN
x = fill(Inf, 10^8)
@time replace!(x, Inf=>NaN)
x = fill(Inf, 10^8)
@time inf2nan(x)
end
bench (generic function with 1 method)
julia> bench()
0.980434 seconds (9 allocations: 774.865 MiB, 0.16% gc time)
0.183578 seconds
0.109929 seconds
julia> bench()
0.971408 seconds (9 allocations: 774.865 MiB, 0.03% gc time)
0.184163 seconds
0.102161 seconds
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
As an additional comment. A standard function to perform this action is replace!. You can use it like this:
julia> a = [1 2 3; 4 5 Inf]
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 Inf
julia> replace!(a, Inf=>NaN)
2×3 Array{Float64,2}:
1.0 2.0 3.0
4.0 5.0 NaN
It will perform better than broadcasting for large arrays.
If you really need speed you can write a simple function like this:
function inf2nan(x)
for i in eachindex(x)
@inbounds x[i] = ifelse(isinf(x[i]), NaN, x[i])
end
end
Now let us simply compare the performance of the three options:
julia> function bench()
x = fill(Inf, 10^8)
@time x[isinf.(x)] .= NaN
x = fill(Inf, 10^8)
@time replace!(x, Inf=>NaN)
x = fill(Inf, 10^8)
@time inf2nan(x)
end
bench (generic function with 1 method)
julia> bench()
0.980434 seconds (9 allocations: 774.865 MiB, 0.16% gc time)
0.183578 seconds
0.109929 seconds
julia> bench()
0.971408 seconds (9 allocations: 774.865 MiB, 0.03% gc time)
0.184163 seconds
0.102161 seconds
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
answered Nov 25 '18 at 7:57
Bogumił KamińskiBogumił Kamiński
13.8k11220
13.8k11220
This is a very useful answer. I typed my answer without thinking about the fact thatisinf.(x)would allocate.
– Colin T Bowers
Nov 25 '18 at 8:40
2
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster thanreplace!in this case because we can avoid branching in the loop usingifelseand since the operation we perform is simple and@inboundsthe compiler can significantly optimize its execution time.
– Bogumił Kamiński
Nov 25 '18 at 8:50
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know whyisinfdoesn't work on arrays anymore.
– Colin T Bowers
Nov 25 '18 at 11:02
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
add a comment |
This is a very useful answer. I typed my answer without thinking about the fact thatisinf.(x)would allocate.
– Colin T Bowers
Nov 25 '18 at 8:40
2
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster thanreplace!in this case because we can avoid branching in the loop usingifelseand since the operation we perform is simple and@inboundsthe compiler can significantly optimize its execution time.
– Bogumił Kamiński
Nov 25 '18 at 8:50
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know whyisinfdoesn't work on arrays anymore.
– Colin T Bowers
Nov 25 '18 at 11:02
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
This is a very useful answer. I typed my answer without thinking about the fact that
isinf.(x) would allocate.– Colin T Bowers
Nov 25 '18 at 8:40
This is a very useful answer. I typed my answer without thinking about the fact that
isinf.(x) would allocate.– Colin T Bowers
Nov 25 '18 at 8:40
2
2
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster than
replace! in this case because we can avoid branching in the loop using ifelse and since the operation we perform is simple and @inbounds the compiler can significantly optimize its execution time.– Bogumił Kamiński
Nov 25 '18 at 8:50
Yes, in Julia - opposite do e.g. R - if you need performance you usually will want to avoid allocation of a binary vector to perform subsetting. In this particular case the loop is faster than
replace! in this case because we can avoid branching in the loop using ifelse and since the operation we perform is simple and @inbounds the compiler can significantly optimize its execution time.– Bogumił Kamiński
Nov 25 '18 at 8:50
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know why
isinf doesn't work on arrays anymore.– Colin T Bowers
Nov 25 '18 at 11:02
Very interesting, thanks for explaining. I'll direct readers to your answer, but leave my answer as it is, since it addresses the part of the question that wanted to know why
isinf doesn't work on arrays anymore.– Colin T Bowers
Nov 25 '18 at 11:02
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
Your answer is the simplest approach in 90% of cases it is good enough and all you need to know, so +1 :).
– Bogumił Kamiński
Nov 25 '18 at 12:58
add a comment |
EDIT: For the most performant approaches to this problem see the excellent answer of @BogumilKaminski. This answer addresses the more general question of why isinf and related functions do not work on arrays anymore.
You are running into the more general issue that lots of functions that worked on arrays pre-v1.0 no longer work on arrays in v1.0 because you are supposed to be using broadcasting. The correct solution for v1.0 is:
a[isinf.(a)] .= NaN
I'm actually broadcasting in two places here. Firstly, we broadcast isinf over the array a, but we are also broadcasting the scalar NaN on the RHS to all indexed locations in the array on the LHS via .=. In general, the dot broadcasting notation is incredibly flexible and performant, and one of my favorite features of the latest iteration of Julia.
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
add a comment |
EDIT: For the most performant approaches to this problem see the excellent answer of @BogumilKaminski. This answer addresses the more general question of why isinf and related functions do not work on arrays anymore.
You are running into the more general issue that lots of functions that worked on arrays pre-v1.0 no longer work on arrays in v1.0 because you are supposed to be using broadcasting. The correct solution for v1.0 is:
a[isinf.(a)] .= NaN
I'm actually broadcasting in two places here. Firstly, we broadcast isinf over the array a, but we are also broadcasting the scalar NaN on the RHS to all indexed locations in the array on the LHS via .=. In general, the dot broadcasting notation is incredibly flexible and performant, and one of my favorite features of the latest iteration of Julia.
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
add a comment |
EDIT: For the most performant approaches to this problem see the excellent answer of @BogumilKaminski. This answer addresses the more general question of why isinf and related functions do not work on arrays anymore.
You are running into the more general issue that lots of functions that worked on arrays pre-v1.0 no longer work on arrays in v1.0 because you are supposed to be using broadcasting. The correct solution for v1.0 is:
a[isinf.(a)] .= NaN
I'm actually broadcasting in two places here. Firstly, we broadcast isinf over the array a, but we are also broadcasting the scalar NaN on the RHS to all indexed locations in the array on the LHS via .=. In general, the dot broadcasting notation is incredibly flexible and performant, and one of my favorite features of the latest iteration of Julia.
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
EDIT: For the most performant approaches to this problem see the excellent answer of @BogumilKaminski. This answer addresses the more general question of why isinf and related functions do not work on arrays anymore.
You are running into the more general issue that lots of functions that worked on arrays pre-v1.0 no longer work on arrays in v1.0 because you are supposed to be using broadcasting. The correct solution for v1.0 is:
a[isinf.(a)] .= NaN
I'm actually broadcasting in two places here. Firstly, we broadcast isinf over the array a, but we are also broadcasting the scalar NaN on the RHS to all indexed locations in the array on the LHS via .=. In general, the dot broadcasting notation is incredibly flexible and performant, and one of my favorite features of the latest iteration of Julia.
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
edited Nov 26 '18 at 21:35
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
answered Nov 25 '18 at 0:35
Colin T BowersColin T Bowers
10.3k43764
10.3k43764
add a comment |
add a comment |
You are passing your entire array to isinf, it doesn't work on arrays, it works on numbers. Try this:
[isinf(i) ? NaN : i for i in a]
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
2
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
add a comment |
You are passing your entire array to isinf, it doesn't work on arrays, it works on numbers. Try this:
[isinf(i) ? NaN : i for i in a]
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
2
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
add a comment |
You are passing your entire array to isinf, it doesn't work on arrays, it works on numbers. Try this:
[isinf(i) ? NaN : i for i in a]
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
You are passing your entire array to isinf, it doesn't work on arrays, it works on numbers. Try this:
[isinf(i) ? NaN : i for i in a]
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
answered Nov 25 '18 at 0:14
chasmanichasmani
1,03211020
1,03211020
2
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
add a comment |
2
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
2
2
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
This actually creates a new array rather than updating the current array. See my answer for more detail.
– Colin T Bowers
Nov 25 '18 at 0:39
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
Yeah fair enough, your answer is better
– chasmani
Nov 26 '18 at 18:02
add a comment |
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