Rank Transformation of an Array
Is there a built in function which rank transforms an array of data? By rank transformation I mean
data = {2.4,5,1,6,7,10,2}
Rank[data]={3,4,1,5,6,7,2}
where each value in data is assigned a rank from minimum to maximum where the lowest value in data is assigned the value of 1, the next highest value is assigned the value of 2, ect.
Ordering does not accomplish this as we obtain
Ordering[data]
{3,7,1,2,4,5,6}
Edit 1: As Carl pointed out, I need to express what I want to happen in the case of a tied ranking. Ultimately, I want to use this rank transformation in the context of the definition of Spearman's Rho function where
Covariance[Transpose[{Rank[X],Rank[Y]}]/(
StandardDeviation[Rank[X]]*StandardDeviation[Rank[Y]])
should equal
SpearmanRho[Transpose[{X,Y}]][[1,2]]
where X and Y are equally lengthed arrays of data.
functions data
asked Nov 26 '18 at 18:37
tquarton
25717
add a comment |
Is there a built in function which rank transforms an array of data? By rank transformation I mean
data = {2.4,5,1,6,7,10,2}
Rank[data]={3,4,1,5,6,7,2}
where each value in data is assigned a rank from minimum to maximum where the lowest value in data is assigned the value of 1, the next highest value is assigned the value of 2, ect.
Ordering does not accomplish this as we obtain
Ordering[data]
{3,7,1,2,4,5,6}
Edit 1: As Carl pointed out, I need to express what I want to happen in the case of a tied ranking. Ultimately, I want to use this rank transformation in the context of the definition of Spearman's Rho function where
Covariance[Transpose[{Rank[X],Rank[Y]}]/(
StandardDeviation[Rank[X]]*StandardDeviation[Rank[Y]])
should equal
SpearmanRho[Transpose[{X,Y}]][[1,2]]
where X and Y are equally lengthed arrays of data.
functions data
asked Nov 26 '18 at 18:37
tquarton
25717
What do you want to return when there are ties?
– Carl Woll
Nov 26 '18 at 19:20
Ah, great question. Give me a moment to respond in this comment with an edit.
– tquarton
Nov 26 '18 at 19:30
I've actually edited the question to address your point Carl.
– tquarton
Nov 26 '18 at 19:38
closely related / possible duplicate: How to get the ranked order
– kglr
Nov 26 '18 at 22:40
add a comment |
Is there a built in function which rank transforms an array of data? By rank transformation I mean
data = {2.4,5,1,6,7,10,2}
Rank[data]={3,4,1,5,6,7,2}
where each value in data is assigned a rank from minimum to maximum where the lowest value in data is assigned the value of 1, the next highest value is assigned the value of 2, ect.
Ordering does not accomplish this as we obtain
Ordering[data]
{3,7,1,2,4,5,6}
Edit 1: As Carl pointed out, I need to express what I want to happen in the case of a tied ranking. Ultimately, I want to use this rank transformation in the context of the definition of Spearman's Rho function where
Covariance[Transpose[{Rank[X],Rank[Y]}]/(
StandardDeviation[Rank[X]]*StandardDeviation[Rank[Y]])
should equal
SpearmanRho[Transpose[{X,Y}]][[1,2]]
where X and Y are equally lengthed arrays of data.
functions data
asked Nov 26 '18 at 18:37
tquarton
25717
Is there a built in function which rank transforms an array of data? By rank transformation I mean
data = {2.4,5,1,6,7,10,2}
Rank[data]={3,4,1,5,6,7,2}
where each value in data is assigned a rank from minimum to maximum where the lowest value in data is assigned the value of 1, the next highest value is assigned the value of 2, ect.
Ordering does not accomplish this as we obtain
Ordering[data]
{3,7,1,2,4,5,6}
Edit 1: As Carl pointed out, I need to express what I want to happen in the case of a tied ranking. Ultimately, I want to use this rank transformation in the context of the definition of Spearman's Rho function where
Covariance[Transpose[{Rank[X],Rank[Y]}]/(
StandardDeviation[Rank[X]]*StandardDeviation[Rank[Y]])
should equal
SpearmanRho[Transpose[{X,Y}]][[1,2]]
where X and Y are equally lengthed arrays of data.
functions data
functions data
asked Nov 26 '18 at 18:37
tquarton
25717
asked Nov 26 '18 at 18:37
tquarton
25717
edited Nov 26 '18 at 19:43
asked Nov 26 '18 at 18:37
tquarton
25717
asked Nov 26 '18 at 18:37
tquarton
25717
asked Nov 26 '18 at 18:37
tquarton
25717
25717
What do you want to return when there are ties?
– Carl Woll
Nov 26 '18 at 19:20
Ah, great question. Give me a moment to respond in this comment with an edit.
– tquarton
Nov 26 '18 at 19:30
I've actually edited the question to address your point Carl.
– tquarton
Nov 26 '18 at 19:38
closely related / possible duplicate: How to get the ranked order
– kglr
Nov 26 '18 at 22:40
add a comment |
What do you want to return when there are ties?
– Carl Woll
Nov 26 '18 at 19:20
Ah, great question. Give me a moment to respond in this comment with an edit.
– tquarton
Nov 26 '18 at 19:30
I've actually edited the question to address your point Carl.
– tquarton
Nov 26 '18 at 19:38
closely related / possible duplicate: How to get the ranked order
– kglr
Nov 26 '18 at 22:40
What do you want to return when there are ties?
– Carl Woll
Nov 26 '18 at 19:20
What do you want to return when there are ties?
– Carl Woll
Nov 26 '18 at 19:20
Ah, great question. Give me a moment to respond in this comment with an edit.
– tquarton
Nov 26 '18 at 19:30
Ah, great question. Give me a moment to respond in this comment with an edit.
– tquarton
Nov 26 '18 at 19:30
I've actually edited the question to address your point Carl.
– tquarton
Nov 26 '18 at 19:38
I've actually edited the question to address your point Carl.
– tquarton
Nov 26 '18 at 19:38
closely related / possible duplicate: How to get the ranked order
– kglr
Nov 26 '18 at 22:40
closely related / possible duplicate: How to get the ranked order
– kglr
Nov 26 '18 at 22:40
add a comment |
3 Answers
3
active
oldest
votes
Statistics`Library`GetDataRankings[{2.4, 5, 1, 6, 7, 10, 2}]
{3, 4, 1, 5, 6, 7, 2}
This gives the same result as Ordering@Ordering@#& if there are no ties in the input data.
If input data has ties:
Statistics`Library`GetDataRankings[{1, 2, 2, 2, 2, 3, 3, 3, 4, 5}]
{1, 7/2, 7/2, 7/2, 7/2, 7, 7, 7, 9, 10}
It is faster than Ordering@Ordering@#& but slower than Henrik Schumacher's Ranking:
SeedRandom[1]
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
0.18
b = Ordering[Ordering[data]]; // RepeatedTiming // First
0.307
c = Statistics`Library`GetDataRankings[data]; // RepeatedTiming // First
0.226
a == b == c
True
A slightly faster alternative (still slower than Ranking):
ranks = Module[{r = Range@Length@#, o = Ordering@#}, Permute[r, o]] &;
d = ranks @ data; // RepeatedTiming // First
0.203
a == b == c == d
True
answered Nov 27 '18 at 0:24
kglr
177k9198404
add a comment |
What about this?
Ordering[Ordering[data]]
{3, 4, 1, 5, 6, 7, 2}
Since Ordering is the bottleneck, here a variant that needs only one call to Ordering:
Ranking[data_] := Module[{a},
a = Range[Length[data]];
a[[Ordering[data]]] = a;
a
]
Comparison:
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
b = Ordering[Ordering[data]]; // RepeatedTiming // First
a == b
0.13
0.234
True
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
add a comment |
I'll answer my own question with a constructed function which does the job:
Rank[x_]:=Flatten[Table[Position[Sort[x], x[[i]]], {i, 1, Length[x]}]]
Ordering gives the sort of inverse of the above function where you get the position of the unsorted data with respect to the sorted data. Here, the Rank function gets the position of the sorted data with respect to the unsorted data.
answered Nov 26 '18 at 18:58
tquarton
25717
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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oldest
votes
Statistics`Library`GetDataRankings[{2.4, 5, 1, 6, 7, 10, 2}]
{3, 4, 1, 5, 6, 7, 2}
This gives the same result as Ordering@Ordering@#& if there are no ties in the input data.
If input data has ties:
Statistics`Library`GetDataRankings[{1, 2, 2, 2, 2, 3, 3, 3, 4, 5}]
{1, 7/2, 7/2, 7/2, 7/2, 7, 7, 7, 9, 10}
It is faster than Ordering@Ordering@#& but slower than Henrik Schumacher's Ranking:
SeedRandom[1]
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
0.18
b = Ordering[Ordering[data]]; // RepeatedTiming // First
0.307
c = Statistics`Library`GetDataRankings[data]; // RepeatedTiming // First
0.226
a == b == c
True
A slightly faster alternative (still slower than Ranking):
ranks = Module[{r = Range@Length@#, o = Ordering@#}, Permute[r, o]] &;
d = ranks @ data; // RepeatedTiming // First
0.203
a == b == c == d
True
answered Nov 27 '18 at 0:24
kglr
177k9198404
add a comment |
Statistics`Library`GetDataRankings[{2.4, 5, 1, 6, 7, 10, 2}]
{3, 4, 1, 5, 6, 7, 2}
This gives the same result as Ordering@Ordering@#& if there are no ties in the input data.
If input data has ties:
Statistics`Library`GetDataRankings[{1, 2, 2, 2, 2, 3, 3, 3, 4, 5}]
{1, 7/2, 7/2, 7/2, 7/2, 7, 7, 7, 9, 10}
It is faster than Ordering@Ordering@#& but slower than Henrik Schumacher's Ranking:
SeedRandom[1]
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
0.18
b = Ordering[Ordering[data]]; // RepeatedTiming // First
0.307
c = Statistics`Library`GetDataRankings[data]; // RepeatedTiming // First
0.226
a == b == c
True
A slightly faster alternative (still slower than Ranking):
ranks = Module[{r = Range@Length@#, o = Ordering@#}, Permute[r, o]] &;
d = ranks @ data; // RepeatedTiming // First
0.203
a == b == c == d
True
answered Nov 27 '18 at 0:24
kglr
177k9198404
add a comment |
Statistics`Library`GetDataRankings[{2.4, 5, 1, 6, 7, 10, 2}]
{3, 4, 1, 5, 6, 7, 2}
This gives the same result as Ordering@Ordering@#& if there are no ties in the input data.
If input data has ties:
Statistics`Library`GetDataRankings[{1, 2, 2, 2, 2, 3, 3, 3, 4, 5}]
{1, 7/2, 7/2, 7/2, 7/2, 7, 7, 7, 9, 10}
It is faster than Ordering@Ordering@#& but slower than Henrik Schumacher's Ranking:
SeedRandom[1]
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
0.18
b = Ordering[Ordering[data]]; // RepeatedTiming // First
0.307
c = Statistics`Library`GetDataRankings[data]; // RepeatedTiming // First
0.226
a == b == c
True
A slightly faster alternative (still slower than Ranking):
ranks = Module[{r = Range@Length@#, o = Ordering@#}, Permute[r, o]] &;
d = ranks @ data; // RepeatedTiming // First
0.203
a == b == c == d
True
answered Nov 27 '18 at 0:24
kglr
177k9198404
Statistics`Library`GetDataRankings[{2.4, 5, 1, 6, 7, 10, 2}]
{3, 4, 1, 5, 6, 7, 2}
This gives the same result as Ordering@Ordering@#& if there are no ties in the input data.
If input data has ties:
Statistics`Library`GetDataRankings[{1, 2, 2, 2, 2, 3, 3, 3, 4, 5}]
{1, 7/2, 7/2, 7/2, 7/2, 7, 7, 7, 9, 10}
It is faster than Ordering@Ordering@#& but slower than Henrik Schumacher's Ranking:
SeedRandom[1]
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
0.18
b = Ordering[Ordering[data]]; // RepeatedTiming // First
0.307
c = Statistics`Library`GetDataRankings[data]; // RepeatedTiming // First
0.226
a == b == c
True
A slightly faster alternative (still slower than Ranking):
ranks = Module[{r = Range@Length@#, o = Ordering@#}, Permute[r, o]] &;
d = ranks @ data; // RepeatedTiming // First
0.203
a == b == c == d
True
answered Nov 27 '18 at 0:24
kglr
177k9198404
edited Nov 27 '18 at 0:31
answered Nov 27 '18 at 0:24
kglr
177k9198404
answered Nov 27 '18 at 0:24
kglr
177k9198404
answered Nov 27 '18 at 0:24
kglr
177k9198404
177k9198404
add a comment |
add a comment |
What about this?
Ordering[Ordering[data]]
{3, 4, 1, 5, 6, 7, 2}
Since Ordering is the bottleneck, here a variant that needs only one call to Ordering:
Ranking[data_] := Module[{a},
a = Range[Length[data]];
a[[Ordering[data]]] = a;
a
]
Comparison:
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
b = Ordering[Ordering[data]]; // RepeatedTiming // First
a == b
0.13
0.234
True
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
add a comment |
What about this?
Ordering[Ordering[data]]
{3, 4, 1, 5, 6, 7, 2}
Since Ordering is the bottleneck, here a variant that needs only one call to Ordering:
Ranking[data_] := Module[{a},
a = Range[Length[data]];
a[[Ordering[data]]] = a;
a
]
Comparison:
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
b = Ordering[Ordering[data]]; // RepeatedTiming // First
a == b
0.13
0.234
True
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
add a comment |
What about this?
Ordering[Ordering[data]]
{3, 4, 1, 5, 6, 7, 2}
Since Ordering is the bottleneck, here a variant that needs only one call to Ordering:
Ranking[data_] := Module[{a},
a = Range[Length[data]];
a[[Ordering[data]]] = a;
a
]
Comparison:
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
b = Ordering[Ordering[data]]; // RepeatedTiming // First
a == b
0.13
0.234
True
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
What about this?
Ordering[Ordering[data]]
{3, 4, 1, 5, 6, 7, 2}
Since Ordering is the bottleneck, here a variant that needs only one call to Ordering:
Ranking[data_] := Module[{a},
a = Range[Length[data]];
a[[Ordering[data]]] = a;
a
]
Comparison:
data = RandomReal[{-1, 1}, 1000000];
a = Ranking[data]; // RepeatedTiming // First
b = Ordering[Ordering[data]]; // RepeatedTiming // First
a == b
0.13
0.234
True
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
edited Nov 26 '18 at 19:18
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
answered Nov 26 '18 at 19:09
Henrik Schumacher
48.8k467139
48.8k467139
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
add a comment |
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
Brilliant! This does it. Thanks very much.
– tquarton
Nov 26 '18 at 19:13
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
You're welcome.
– Henrik Schumacher
Nov 26 '18 at 19:14
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
Carl Woll brought up a great point with regards to tied rankings. I thought I'd ping you in this comment in case you wanted to address it. For my purposes, I don't believe there are ties in my dataset of interest, so your solution still holds.
– tquarton
Nov 26 '18 at 19:37
add a comment |
I'll answer my own question with a constructed function which does the job:
Rank[x_]:=Flatten[Table[Position[Sort[x], x[[i]]], {i, 1, Length[x]}]]
Ordering gives the sort of inverse of the above function where you get the position of the unsorted data with respect to the sorted data. Here, the Rank function gets the position of the sorted data with respect to the unsorted data.
answered Nov 26 '18 at 18:58
tquarton
25717
add a comment |
I'll answer my own question with a constructed function which does the job:
Rank[x_]:=Flatten[Table[Position[Sort[x], x[[i]]], {i, 1, Length[x]}]]
Ordering gives the sort of inverse of the above function where you get the position of the unsorted data with respect to the sorted data. Here, the Rank function gets the position of the sorted data with respect to the unsorted data.
answered Nov 26 '18 at 18:58
tquarton
25717
add a comment |
I'll answer my own question with a constructed function which does the job:
Rank[x_]:=Flatten[Table[Position[Sort[x], x[[i]]], {i, 1, Length[x]}]]
Ordering gives the sort of inverse of the above function where you get the position of the unsorted data with respect to the sorted data. Here, the Rank function gets the position of the sorted data with respect to the unsorted data.
answered Nov 26 '18 at 18:58
tquarton
25717
I'll answer my own question with a constructed function which does the job:
Rank[x_]:=Flatten[Table[Position[Sort[x], x[[i]]], {i, 1, Length[x]}]]
Ordering gives the sort of inverse of the above function where you get the position of the unsorted data with respect to the sorted data. Here, the Rank function gets the position of the sorted data with respect to the unsorted data.
answered Nov 26 '18 at 18:58
tquarton
25717
edited Nov 26 '18 at 19:07
answered Nov 26 '18 at 18:58
tquarton
25717
answered Nov 26 '18 at 18:58
tquarton
25717
answered Nov 26 '18 at 18:58
tquarton
25717
25717
add a comment |
add a comment |
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What do you want to return when there are ties?
– Carl Woll
Nov 26 '18 at 19:20
Ah, great question. Give me a moment to respond in this comment with an edit.
– tquarton
Nov 26 '18 at 19:30
I've actually edited the question to address your point Carl.
– tquarton
Nov 26 '18 at 19:38
closely related / possible duplicate: How to get the ranked order
– kglr
Nov 26 '18 at 22:40