if I have a list of numbers and I want to know where a number lies between two numbers in that list.

2 comentarios

Roger Stafford
Roger Stafford el 19 de Dic. de 2016
Please give more details in your question! Better still, give a concrete example of the "list" and precisely what results you wish to obtain from it.
khamiis E
khamiis E el 19 de Dic. de 2016

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Guillaume
Guillaume el 19 de Dic. de 2016

1 voto

A = [10 25
30 45
50 150
300 450
500 501
502 600
630 700
720 800
801 815
820 1000];
B = 33;
find(B >= A(:, 1) & B <= A(:, 2))

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Walter Roberson
Walter Roberson el 19 de Dic. de 2016

1 voto

Suppose you have a vector of values in sorted order, and have some other values, and you want to know where in the vector the other values would sit. Then:
[~, ~, binnumber] = histcounts(TheOtherValues, TheVectorOfFixedValues);
Then, TheOtherValues(K) is between TheVectorOfFixedValues(binnumber(K)) and the next value.

5 comentarios

khamiis E
khamiis E el 19 de Dic. de 2016
Editada: khamiis E el 19 de Dic. de 2016
Thanks alot for your answer. I can't find the histcounts() function. Is there alternative function?
Star Strider
Star Strider el 19 de Dic. de 2016
The previous histcounts function was called histc with a slightly different calling and output syntax.
Walter Roberson
Walter Roberson el 19 de Dic. de 2016
[~, binnumber] = histc(TheOtherValues, TheVectorOfFixedValues);
Walter Roberson
Walter Roberson el 19 de Dic. de 2016
Note: this code was written assuming that your ranges had no gaps, but it appears from your diagram that it does not apply.
khamiis E
khamiis E el 20 de Dic. de 2016
I really appreciate your effort . thanks for your time and your answer

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Andrei Bobrov
Andrei Bobrov el 19 de Dic. de 2016
Editada: Andrei Bobrov el 19 de Dic. de 2016

1 voto

A = [10 25
30 45
50 150
300 450
500 501
502 600
630 700
720 800
801 815
820 1000]; % your "start-end"
B = [33 300 501 75 754 809 1000 47]'; % Let B - your numbers
b = prod(A - reshape(B,1,1,[]),2);
out = sum(bsxfun(@times,squeeze(b < 0 | b == 0),(1:size(A,1))'))

1 comentario

khamiis E
khamiis E el 20 de Dic. de 2016
I really appreciate your effort . Thanks for your answer.

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