How to creates a Counts[] function that includes zero?
Clash Royale CLAN TAG#URR8PPP
In the following example, I'm trying to how many times each integer 1-n appears in list 'a'. But if an element did not appear in a list, Counts creates an association that returns the element instead of zero. So the result is a mixed list of counts and integers, with no indication of which you're looking at.
n = 10;
a = RandomInteger[n, n]
c = Counts[a];
Table[i /. c, i, 1, n]
7, 7, 0, 7, 10, 5, 3, 0, 1, 8
1, 2, 1, 4, 1, 6, 3, 1, 9, 1
Is there a way to make a Counts function that can count zero?
list-manipulation counting
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
asked Dec 13 at 21:34
Jerry Guern
1,983833
add a comment |
In the following example, I'm trying to how many times each integer 1-n appears in list 'a'. But if an element did not appear in a list, Counts creates an association that returns the element instead of zero. So the result is a mixed list of counts and integers, with no indication of which you're looking at.
n = 10;
a = RandomInteger[n, n]
c = Counts[a];
Table[i /. c, i, 1, n]
7, 7, 0, 7, 10, 5, 3, 0, 1, 8
1, 2, 1, 4, 1, 6, 3, 1, 9, 1
Is there a way to make a Counts function that can count zero?
list-manipulation counting
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
asked Dec 13 at 21:34
Jerry Guern
1,983833
add a comment |
In the following example, I'm trying to how many times each integer 1-n appears in list 'a'. But if an element did not appear in a list, Counts creates an association that returns the element instead of zero. So the result is a mixed list of counts and integers, with no indication of which you're looking at.
n = 10;
a = RandomInteger[n, n]
c = Counts[a];
Table[i /. c, i, 1, n]
7, 7, 0, 7, 10, 5, 3, 0, 1, 8
1, 2, 1, 4, 1, 6, 3, 1, 9, 1
Is there a way to make a Counts function that can count zero?
list-manipulation counting
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
asked Dec 13 at 21:34
Jerry Guern
1,983833
In the following example, I'm trying to how many times each integer 1-n appears in list 'a'. But if an element did not appear in a list, Counts creates an association that returns the element instead of zero. So the result is a mixed list of counts and integers, with no indication of which you're looking at.
n = 10;
a = RandomInteger[n, n]
c = Counts[a];
Table[i /. c, i, 1, n]
7, 7, 0, 7, 10, 5, 3, 0, 1, 8
1, 2, 1, 4, 1, 6, 3, 1, 9, 1
Is there a way to make a Counts function that can count zero?
list-manipulation counting
list-manipulation counting
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
asked Dec 13 at 21:34
Jerry Guern
1,983833
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
asked Dec 13 at 21:34
Jerry Guern
1,983833
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
edited Dec 13 at 21:50
Henrik Schumacher
48.1k467136
48.1k467136
asked Dec 13 at 21:34
Jerry Guern
1,983833
asked Dec 13 at 21:34
Jerry Guern
1,983833
asked Dec 13 at 21:34
Jerry Guern
1,983833
1,983833
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Use Lookup instead of ReplaceAll:
n = 10;
SeedRandom[1]
a = RandomInteger[n,n];
c = Counts[a]
Table[i /. c, i, 1, n]
Lookup[c, Range[10], 0]
<|1 -> 1, 4 -> 2, 0 -> 4, 7 -> 1, 8 -> 1, 6 -> 1|>
1, 2, 3, 2, 5, 1, 1, 1, 9, 10
1, 0, 0, 2, 0, 1, 1, 1, 0, 0
answered Dec 13 at 21:38
Carl Woll
66.8k386174
add a comment |
Or you can use SparseArray with additive matrix assembly which may be a bit faster:
n = 1000000;
a = RandomInteger[n, n];
counts = Lookup[Counts[a], Range[1, n], 0]; // AbsoluteTiming // First
counts2 = Rest@With[spopt = SystemOptions["SparseArrayOptions"],
Internal`WithLocalSettings[
SetSystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries" -> Total],
SparseArray[Partition[a+1, 1] -> 1, n+1],
SetSystemOptions[spopt]]
]; // AbsoluteTiming // First
1.30965
0.148508
True
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
add a comment |
a = 7, 7, 0, 7, 10, 5, 3, 0, 1, 8;
Counts[a] /@ Range[10] /. Missing -> (0 &)
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
Alternatively,
Block[Missing = (0 &), Counts[a] /@ Range[10]]
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
answered Dec 13 at 22:10
kglr
176k9197402
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
Use Lookup instead of ReplaceAll:
n = 10;
SeedRandom[1]
a = RandomInteger[n,n];
c = Counts[a]
Table[i /. c, i, 1, n]
Lookup[c, Range[10], 0]
<|1 -> 1, 4 -> 2, 0 -> 4, 7 -> 1, 8 -> 1, 6 -> 1|>
1, 2, 3, 2, 5, 1, 1, 1, 9, 10
1, 0, 0, 2, 0, 1, 1, 1, 0, 0
answered Dec 13 at 21:38
Carl Woll
66.8k386174
add a comment |
Use Lookup instead of ReplaceAll:
n = 10;
SeedRandom[1]
a = RandomInteger[n,n];
c = Counts[a]
Table[i /. c, i, 1, n]
Lookup[c, Range[10], 0]
<|1 -> 1, 4 -> 2, 0 -> 4, 7 -> 1, 8 -> 1, 6 -> 1|>
1, 2, 3, 2, 5, 1, 1, 1, 9, 10
1, 0, 0, 2, 0, 1, 1, 1, 0, 0
answered Dec 13 at 21:38
Carl Woll
66.8k386174
add a comment |
Use Lookup instead of ReplaceAll:
n = 10;
SeedRandom[1]
a = RandomInteger[n,n];
c = Counts[a]
Table[i /. c, i, 1, n]
Lookup[c, Range[10], 0]
<|1 -> 1, 4 -> 2, 0 -> 4, 7 -> 1, 8 -> 1, 6 -> 1|>
1, 2, 3, 2, 5, 1, 1, 1, 9, 10
1, 0, 0, 2, 0, 1, 1, 1, 0, 0
answered Dec 13 at 21:38
Carl Woll
66.8k386174
Use Lookup instead of ReplaceAll:
n = 10;
SeedRandom[1]
a = RandomInteger[n,n];
c = Counts[a]
Table[i /. c, i, 1, n]
Lookup[c, Range[10], 0]
<|1 -> 1, 4 -> 2, 0 -> 4, 7 -> 1, 8 -> 1, 6 -> 1|>
1, 2, 3, 2, 5, 1, 1, 1, 9, 10
1, 0, 0, 2, 0, 1, 1, 1, 0, 0
answered Dec 13 at 21:38
Carl Woll
66.8k386174
answered Dec 13 at 21:38
Carl Woll
66.8k386174
answered Dec 13 at 21:38
Carl Woll
66.8k386174
answered Dec 13 at 21:38
Carl Woll
66.8k386174
66.8k386174
add a comment |
add a comment |
Or you can use SparseArray with additive matrix assembly which may be a bit faster:
n = 1000000;
a = RandomInteger[n, n];
counts = Lookup[Counts[a], Range[1, n], 0]; // AbsoluteTiming // First
counts2 = Rest@With[spopt = SystemOptions["SparseArrayOptions"],
Internal`WithLocalSettings[
SetSystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries" -> Total],
SparseArray[Partition[a+1, 1] -> 1, n+1],
SetSystemOptions[spopt]]
]; // AbsoluteTiming // First
1.30965
0.148508
True
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
add a comment |
Or you can use SparseArray with additive matrix assembly which may be a bit faster:
n = 1000000;
a = RandomInteger[n, n];
counts = Lookup[Counts[a], Range[1, n], 0]; // AbsoluteTiming // First
counts2 = Rest@With[spopt = SystemOptions["SparseArrayOptions"],
Internal`WithLocalSettings[
SetSystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries" -> Total],
SparseArray[Partition[a+1, 1] -> 1, n+1],
SetSystemOptions[spopt]]
]; // AbsoluteTiming // First
1.30965
0.148508
True
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
add a comment |
Or you can use SparseArray with additive matrix assembly which may be a bit faster:
n = 1000000;
a = RandomInteger[n, n];
counts = Lookup[Counts[a], Range[1, n], 0]; // AbsoluteTiming // First
counts2 = Rest@With[spopt = SystemOptions["SparseArrayOptions"],
Internal`WithLocalSettings[
SetSystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries" -> Total],
SparseArray[Partition[a+1, 1] -> 1, n+1],
SetSystemOptions[spopt]]
]; // AbsoluteTiming // First
1.30965
0.148508
True
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
Or you can use SparseArray with additive matrix assembly which may be a bit faster:
n = 1000000;
a = RandomInteger[n, n];
counts = Lookup[Counts[a], Range[1, n], 0]; // AbsoluteTiming // First
counts2 = Rest@With[spopt = SystemOptions["SparseArrayOptions"],
Internal`WithLocalSettings[
SetSystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries" -> Total],
SparseArray[Partition[a+1, 1] -> 1, n+1],
SetSystemOptions[spopt]]
]; // AbsoluteTiming // First
1.30965
0.148508
True
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
answered Dec 13 at 21:49
Henrik Schumacher
48.1k467136
48.1k467136
add a comment |
add a comment |
a = 7, 7, 0, 7, 10, 5, 3, 0, 1, 8;
Counts[a] /@ Range[10] /. Missing -> (0 &)
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
Alternatively,
Block[Missing = (0 &), Counts[a] /@ Range[10]]
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
answered Dec 13 at 22:10
kglr
176k9197402
add a comment |
a = 7, 7, 0, 7, 10, 5, 3, 0, 1, 8;
Counts[a] /@ Range[10] /. Missing -> (0 &)
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
Alternatively,
Block[Missing = (0 &), Counts[a] /@ Range[10]]
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
answered Dec 13 at 22:10
kglr
176k9197402
add a comment |
a = 7, 7, 0, 7, 10, 5, 3, 0, 1, 8;
Counts[a] /@ Range[10] /. Missing -> (0 &)
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
Alternatively,
Block[Missing = (0 &), Counts[a] /@ Range[10]]
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
answered Dec 13 at 22:10
kglr
176k9197402
a = 7, 7, 0, 7, 10, 5, 3, 0, 1, 8;
Counts[a] /@ Range[10] /. Missing -> (0 &)
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
Alternatively,
Block[Missing = (0 &), Counts[a] /@ Range[10]]
1, 0, 1, 0, 1, 0, 3, 1, 0, 1
answered Dec 13 at 22:10
kglr
176k9197402
answered Dec 13 at 22:10
kglr
176k9197402
answered Dec 13 at 22:10
kglr
176k9197402
answered Dec 13 at 22:10
kglr
176k9197402
176k9197402
add a comment |
add a comment |
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