How to creates a Counts[] function that includes zero?
up vote
2
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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 2 hours ago
Henrik Schumacher
47.3k466134
asked 3 hours ago
Jerry Guern
1,976833
add a comment |
up vote
2
down vote
favorite
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 2 hours ago
Henrik Schumacher
47.3k466134
asked 3 hours ago
Jerry Guern
1,976833
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 2 hours ago
Henrik Schumacher
47.3k466134
asked 3 hours ago
Jerry Guern
1,976833
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 2 hours ago
Henrik Schumacher
47.3k466134
asked 3 hours ago
Jerry Guern
1,976833
edited 2 hours ago
Henrik Schumacher
47.3k466134
asked 3 hours ago
Jerry Guern
1,976833
edited 2 hours ago
Henrik Schumacher
47.3k466134
edited 2 hours ago
Henrik Schumacher
47.3k466134
edited 2 hours ago
Henrik Schumacher
47.3k466134
47.3k466134
asked 3 hours ago
Jerry Guern
1,976833
asked 3 hours ago
Jerry Guern
1,976833
asked 3 hours ago
Jerry Guern
1,976833
1,976833
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
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 3 hours ago
Carl Woll
66.5k385174
add a comment |
up vote
1
down vote
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 2 hours ago
Henrik Schumacher
47.3k466134
add a comment |
up vote
0
down vote
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 2 hours ago
kglr
175k9197402
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
up vote
3
down vote
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 3 hours ago
Carl Woll
66.5k385174
add a comment |
up vote
3
down vote
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 3 hours ago
Carl Woll
66.5k385174
add a comment |
up vote
3
down vote
up vote
3
down vote
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 3 hours ago
Carl Woll
66.5k385174
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 3 hours ago
Carl Woll
66.5k385174
answered 3 hours ago
Carl Woll
66.5k385174
answered 3 hours ago
Carl Woll
66.5k385174
answered 3 hours ago
Carl Woll
66.5k385174
66.5k385174
add a comment |
add a comment |
up vote
1
down vote
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 2 hours ago
Henrik Schumacher
47.3k466134
add a comment |
up vote
1
down vote
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 2 hours ago
Henrik Schumacher
47.3k466134
add a comment |
up vote
1
down vote
up vote
1
down vote
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 2 hours ago
Henrik Schumacher
47.3k466134
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 2 hours ago
Henrik Schumacher
47.3k466134
answered 2 hours ago
Henrik Schumacher
47.3k466134
answered 2 hours ago
Henrik Schumacher
47.3k466134
answered 2 hours ago
Henrik Schumacher
47.3k466134
47.3k466134
add a comment |
add a comment |
up vote
0
down vote
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 2 hours ago
kglr
175k9197402
add a comment |
up vote
0
down vote
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 2 hours ago
kglr
175k9197402
add a comment |
up vote
0
down vote
up vote
0
down vote
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 2 hours ago
kglr
175k9197402
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 2 hours ago
kglr
175k9197402
answered 2 hours ago
kglr
175k9197402
answered 2 hours ago
kglr
175k9197402
answered 2 hours ago
kglr
175k9197402
175k9197402
add a comment |
add a comment |
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