selecting certain values from a list

2011 6 28
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I have this list below (call it A), and i wish to find the last value of a group of numbers. if you look at the numbers, they are bunched together in groups (not always consecutively), and i only want the last number in each group.
A =
8
9
106
107
109
110
244
245
325
326
329
334
335
483
484
485
486
527
528
690
691
So the new list (call it B), should look like this..
B =
9
110
245
335
486
528
691
I was thinking of making a for loop (i=1:length(A)) and doing a 'find' to see if a number exists in between that number and +20 after it (each new group of numbers never exceeds beyond 20).
is this possible? thank you in advance

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Paulo Silva
Paulo Silva 2011-6-28

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M=[ 8 9 106 107 109 110 244 245 325 326 329 334 335 483 484 485 486 527 528 690 691]
d=diff(M); %find the diference between consecutive values
[M(d>1);M(end)] %get the values with diference bigger than 1
append M(end) because the last one is always the last from a group and the diference function doesn't catch it

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Nathan Greco
Nathan Greco 2011-6-28
Note the comment that says "they are bunched together in groups (not always consecutively)". Yours counts groups as consecutive, while the following is in actuality one group: 325 326 329 334 335.
Oleg Komarov
Oleg Komarov 2011-6-28
Andrei's solution applies well to your example data. The implicit rule is that the group is defined by the hundreds/thousands/so on digits.
Sean de Wolski
Sean de Wolski 2011-6-28
I think Paulo's code mended to be >20 as the OP said would be ideal.
Oleg Komarov
Oleg Komarov 2011-6-28
What if the numbers are: 398 399 400 401. Will the 3.. and the 4.. numbers be part of the same group or to separate?
Sean de Wolski
Sean de Wolski 2011-6-28
OP: "I was thinking of making a for loop (i=1:length(A)) and doing a 'find' to see if a number exists in between that number and +20 after it (each new group of numbers never exceeds beyond 20)."
so basically anywhere there's a gap of more than 20 it fails.
Michael
Michael 2011-6-28
if the numbers are 398 399 400 401, then in my case they will all belong to the same group. these numbers in my case represent indices for a data set, and the gap between groups of data is usually higher than 100, and each different "cluster" of close numbers usually doesnt exceed 10, but i used 20 to be safe, with the intention of only gathering the last value of each group.

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Andrei Bobrov
Andrei Bobrov 2011-6-28

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[a b c]=unique(floor(A/100));
out = A(b)

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Michael
Michael 2011-6-28
nice thinking, this worked, but for some reason when i added the longer list i have, one of them didnt show up..

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