Count values in a vector less than each one of the elements in another vector

I have two vectors r and d and I'd like to know the number of times r<d(i) where i=1:length(d).

r=rand(1,1E7);

d=linspace(0,1,10);

So far I've got the following, but it's not very elegant:

for i=1:length(d)

sum(r<d(i))

end

This is an example in R but I'm not really sure this would work for matlab:
Finding number of elements in one vector that are less than an element in another vector

edited Nov 16 '18 at 16:48

asked Nov 16 '18 at 16:26

HCAI

57541338

add a comment |

I have two vectors r and d and I'd like to know the number of times r<d(i) where i=1:length(d).

r=rand(1,1E7);

d=linspace(0,1,10);

So far I've got the following, but it's not very elegant:

for i=1:length(d)

sum(r<d(i))

end

This is an example in R but I'm not really sure this would work for matlab:
Finding number of elements in one vector that are less than an element in another vector

edited Nov 16 '18 at 16:48

asked Nov 16 '18 at 16:26

HCAI

57541338

add a comment |

I have two vectors r and d and I'd like to know the number of times r<d(i) where i=1:length(d).

r=rand(1,1E7);

d=linspace(0,1,10);

So far I've got the following, but it's not very elegant:

for i=1:length(d)

sum(r<d(i))

end

This is an example in R but I'm not really sure this would work for matlab:
Finding number of elements in one vector that are less than an element in another vector

edited Nov 16 '18 at 16:48

asked Nov 16 '18 at 16:26

HCAI

57541338

I have two vectors r and d and I'd like to know the number of times r<d(i) where i=1:length(d).

r=rand(1,1E7);

d=linspace(0,1,10);

So far I've got the following, but it's not very elegant:

for i=1:length(d)

sum(r<d(i))

end

This is an example in R but I'm not really sure this would work for matlab:
Finding number of elements in one vector that are less than an element in another vector

arrays matlab vector

edited Nov 16 '18 at 16:48

asked Nov 16 '18 at 16:26

HCAI

57541338

edited Nov 16 '18 at 16:48

asked Nov 16 '18 at 16:26

HCAI

57541338

edited Nov 16 '18 at 16:48

asked Nov 16 '18 at 16:26

HCAI

57541338

asked Nov 16 '18 at 16:26

HCAI

57541338

asked Nov 16 '18 at 16:26

HCAI

57541338

add a comment |

3 Answers
3

active

oldest

votes

You can use singleton expansion with bsxfun: faster, more elegant than the loop, but also more memory-intensive:

result = sum(bsxfun(@lt, r(:), d(:).'), 1);

In recent Matlab versions bsxfun can be dropped thanks to implicit singleton expansion:

result = sum(r(:)<d(:).', 1);

An alternative approach is to use the histcounts function with the 'cumcount' option:

result = histcounts(r(:), [-inf; d(:); inf], 'Normalization', 'cumcount');

result = result(1:end-1);

edited Nov 16 '18 at 16:48

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

1

aw, 50 seconds late...

– Brice
Nov 16 '18 at 16:36

@Brice ¯_(ツ)_/¯ :-D

– Luis Mendo
Nov 16 '18 at 16:37

Thank you for such a quick reply! The first one works for me with the example vectors but my actual r is 1E7 in length.... whereas d is only 10. It says i will run out of memory. Shall I update my question to reflect this. I'm sorry for the confusion.

– HCAI
Nov 16 '18 at 16:46

@HCAI The histcounts approach should be better then

– Luis Mendo
Nov 16 '18 at 16:47

1

Thank you very much, I'd like to accept your answer. The histcount is super fast!!!

– HCAI
Nov 16 '18 at 16:57

add a comment |

You may build a matrix flagging values from vector r inferior to values from vector d in one time with bsxfun, then sum the values:

flag=bsxfun(@lt,r',d);

result=sum(flag,1);

answered Nov 16 '18 at 16:35

Brice

1,400110

Thank you for your answer, I'd like to accept Luis' as the histcount seems to be quite a bit quicker although this also works.

– HCAI
Nov 16 '18 at 16:58

add a comment |

For each element in d, count how many times this element is bigger than the elements in r, which is equivalent to your problem.

r=rand(1,10);

d=linspace(0,1,10);



result = sum(d>r(:))

Output:

result =

     0     0     1     2     7     8     8     8     9    10

edited Nov 16 '18 at 16:54

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You can use singleton expansion with bsxfun: faster, more elegant than the loop, but also more memory-intensive:

result = sum(bsxfun(@lt, r(:), d(:).'), 1);

In recent Matlab versions bsxfun can be dropped thanks to implicit singleton expansion:

result = sum(r(:)<d(:).', 1);

An alternative approach is to use the histcounts function with the 'cumcount' option:

result = histcounts(r(:), [-inf; d(:); inf], 'Normalization', 'cumcount');

result = result(1:end-1);

edited Nov 16 '18 at 16:48

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

1

aw, 50 seconds late...

– Brice
Nov 16 '18 at 16:36

@Brice ¯_(ツ)_/¯ :-D

– Luis Mendo
Nov 16 '18 at 16:37

Thank you for such a quick reply! The first one works for me with the example vectors but my actual r is 1E7 in length.... whereas d is only 10. It says i will run out of memory. Shall I update my question to reflect this. I'm sorry for the confusion.

– HCAI
Nov 16 '18 at 16:46

@HCAI The histcounts approach should be better then

– Luis Mendo
Nov 16 '18 at 16:47

1

Thank you very much, I'd like to accept your answer. The histcount is super fast!!!

– HCAI
Nov 16 '18 at 16:57

add a comment |

You can use singleton expansion with bsxfun: faster, more elegant than the loop, but also more memory-intensive:

result = sum(bsxfun(@lt, r(:), d(:).'), 1);

In recent Matlab versions bsxfun can be dropped thanks to implicit singleton expansion:

result = sum(r(:)<d(:).', 1);

An alternative approach is to use the histcounts function with the 'cumcount' option:

result = histcounts(r(:), [-inf; d(:); inf], 'Normalization', 'cumcount');

result = result(1:end-1);

edited Nov 16 '18 at 16:48

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

1

aw, 50 seconds late...

– Brice
Nov 16 '18 at 16:36

@Brice ¯_(ツ)_/¯ :-D

– Luis Mendo
Nov 16 '18 at 16:37

Thank you for such a quick reply! The first one works for me with the example vectors but my actual r is 1E7 in length.... whereas d is only 10. It says i will run out of memory. Shall I update my question to reflect this. I'm sorry for the confusion.

– HCAI
Nov 16 '18 at 16:46

@HCAI The histcounts approach should be better then

– Luis Mendo
Nov 16 '18 at 16:47

1

Thank you very much, I'd like to accept your answer. The histcount is super fast!!!

– HCAI
Nov 16 '18 at 16:57

add a comment |

You can use singleton expansion with bsxfun: faster, more elegant than the loop, but also more memory-intensive:

result = sum(bsxfun(@lt, r(:), d(:).'), 1);

In recent Matlab versions bsxfun can be dropped thanks to implicit singleton expansion:

result = sum(r(:)<d(:).', 1);

An alternative approach is to use the histcounts function with the 'cumcount' option:

result = histcounts(r(:), [-inf; d(:); inf], 'Normalization', 'cumcount');

result = result(1:end-1);

edited Nov 16 '18 at 16:48

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

You can use singleton expansion with bsxfun: faster, more elegant than the loop, but also more memory-intensive:

result = sum(bsxfun(@lt, r(:), d(:).'), 1);

In recent Matlab versions bsxfun can be dropped thanks to implicit singleton expansion:

result = sum(r(:)<d(:).', 1);

An alternative approach is to use the histcounts function with the 'cumcount' option:

result = histcounts(r(:), [-inf; d(:); inf], 'Normalization', 'cumcount');

result = result(1:end-1);

edited Nov 16 '18 at 16:48

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

edited Nov 16 '18 at 16:48

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

answered Nov 16 '18 at 16:34

Luis Mendo

93.2k1154122

1

aw, 50 seconds late...

– Brice
Nov 16 '18 at 16:36

@Brice ¯_(ツ)_/¯ :-D

– Luis Mendo
Nov 16 '18 at 16:37

Thank you for such a quick reply! The first one works for me with the example vectors but my actual r is 1E7 in length.... whereas d is only 10. It says i will run out of memory. Shall I update my question to reflect this. I'm sorry for the confusion.

– HCAI
Nov 16 '18 at 16:46

@HCAI The histcounts approach should be better then

– Luis Mendo
Nov 16 '18 at 16:47

1

Thank you very much, I'd like to accept your answer. The histcount is super fast!!!

– HCAI
Nov 16 '18 at 16:57

add a comment |

1

aw, 50 seconds late...

– Brice
Nov 16 '18 at 16:36

@Brice ¯_(ツ)_/¯ :-D

– Luis Mendo
Nov 16 '18 at 16:37

Thank you for such a quick reply! The first one works for me with the example vectors but my actual r is 1E7 in length.... whereas d is only 10. It says i will run out of memory. Shall I update my question to reflect this. I'm sorry for the confusion.

– HCAI
Nov 16 '18 at 16:46

@HCAI The histcounts approach should be better then

– Luis Mendo
Nov 16 '18 at 16:47

1

Thank you very much, I'd like to accept your answer. The histcount is super fast!!!

– HCAI
Nov 16 '18 at 16:57

aw, 50 seconds late...

– Brice
Nov 16 '18 at 16:36

@Brice ¯_(ツ)_/¯ :-D

– Luis Mendo
Nov 16 '18 at 16:37

Thank you for such a quick reply! The first one works for me with the example vectors but my actual r is 1E7 in length.... whereas d is only 10. It says i will run out of memory. Shall I update my question to reflect this. I'm sorry for the confusion.

– HCAI
Nov 16 '18 at 16:46

@HCAI The histcounts approach should be better then

– Luis Mendo
Nov 16 '18 at 16:47

Thank you very much, I'd like to accept your answer. The histcount is super fast!!!

– HCAI
Nov 16 '18 at 16:57

add a comment |

You may build a matrix flagging values from vector r inferior to values from vector d in one time with bsxfun, then sum the values:

flag=bsxfun(@lt,r',d);

result=sum(flag,1);

answered Nov 16 '18 at 16:35

Brice

1,400110

Thank you for your answer, I'd like to accept Luis' as the histcount seems to be quite a bit quicker although this also works.

– HCAI
Nov 16 '18 at 16:58

add a comment |

You may build a matrix flagging values from vector r inferior to values from vector d in one time with bsxfun, then sum the values:

flag=bsxfun(@lt,r',d);

result=sum(flag,1);

answered Nov 16 '18 at 16:35

Brice

1,400110

Thank you for your answer, I'd like to accept Luis' as the histcount seems to be quite a bit quicker although this also works.

– HCAI
Nov 16 '18 at 16:58

add a comment |

You may build a matrix flagging values from vector r inferior to values from vector d in one time with bsxfun, then sum the values:

flag=bsxfun(@lt,r',d);

result=sum(flag,1);

answered Nov 16 '18 at 16:35

Brice

1,400110

You may build a matrix flagging values from vector r inferior to values from vector d in one time with bsxfun, then sum the values:

flag=bsxfun(@lt,r',d);

result=sum(flag,1);

answered Nov 16 '18 at 16:35

Brice

1,400110

answered Nov 16 '18 at 16:35

Brice

1,400110

answered Nov 16 '18 at 16:35

Brice

1,400110

answered Nov 16 '18 at 16:35

Brice

1,400110

Thank you for your answer, I'd like to accept Luis' as the histcount seems to be quite a bit quicker although this also works.

– HCAI
Nov 16 '18 at 16:58

add a comment |

Thank you for your answer, I'd like to accept Luis' as the histcount seems to be quite a bit quicker although this also works.

– HCAI
Nov 16 '18 at 16:58

Thank you for your answer, I'd like to accept Luis' as the histcount seems to be quite a bit quicker although this also works.

– HCAI
Nov 16 '18 at 16:58

add a comment |

For each element in d, count how many times this element is bigger than the elements in r, which is equivalent to your problem.

r=rand(1,10);

d=linspace(0,1,10);



result = sum(d>r(:))

Output:

result =

     0     0     1     2     7     8     8     8     9    10

edited Nov 16 '18 at 16:54

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

add a comment |

For each element in d, count how many times this element is bigger than the elements in r, which is equivalent to your problem.

r=rand(1,10);

d=linspace(0,1,10);



result = sum(d>r(:))

Output:

result =

     0     0     1     2     7     8     8     8     9    10

edited Nov 16 '18 at 16:54

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

add a comment |

For each element in d, count how many times this element is bigger than the elements in r, which is equivalent to your problem.

r=rand(1,10);

d=linspace(0,1,10);



result = sum(d>r(:))

Output:

result =

     0     0     1     2     7     8     8     8     9    10

edited Nov 16 '18 at 16:54

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

For each element in d, count how many times this element is bigger than the elements in r, which is equivalent to your problem.

r=rand(1,10);

d=linspace(0,1,10);



result = sum(d>r(:))

Output:

result =

     0     0     1     2     7     8     8     8     9    10

edited Nov 16 '18 at 16:54

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

edited Nov 16 '18 at 16:54

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

answered Nov 16 '18 at 16:39

Banghua Zhao

1,2771719

add a comment |

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