# Counting in one column that is conditional on another column

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Mirthand 2021 年 4 月 2 日
コメント済み: Mirthand 2021 年 4 月 5 日
For each cell (or "trial") in the variable trials (attached):
I'd like to count the number of 1's in column 4 that is dependent on certain conditions in column 2 for each trial.
Conditions saved in their own variable:
Count 1's in column 4 if column 2 is between 0 and 4 (less than 4, not equal to)
Count 1's in column 4 if column 2 is between 4 and 5 (less than 5, not equal to)
Count 1's in column 4 if column 2 is between 5 and 6 (less than 6, not equal to)
Count 1's in column 4 if column 2 is between 6 and 7 (less than 7, not equal to)
Count 1's in column 4 if column 2 is between 7 and end

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### 採用された回答

per isakson 2021 年 4 月 2 日

Does this do what you look for?
%%
limits = { [0;4], [4;5], [5;6], [6;7], [7;inf] };
%% some simple test data
trials = {[ 0, 2, 0, 1, 0, 0, 0
0, 2, 0, 0, 0, 0, 0
0, 5, 0, 1, 0, 0, 0
0, 5, 0, 0, 0, 0, 0
0, 5, 0, 1, 0, 0, 0
0, 5, 0, 0, 0, 0, 0
0, 9, 0, 1, 0, 0, 0 ]};
%%
fh = @(trial,lim) sum( (trial(:,4)==1) & (lim(1)<=trial(:,2) & trial(:,2)<lim(2)), 1 );
%%
counts = zeros( numel(trials), numel(limits) );
for rr = 1 : numel(trials)
for cc = 1 : numel(limits)
counts(rr,cc) = fh( trials{rr}, limits{cc} );
end
end
Inspect
>> counts
counts =
1 0 2 0 1
Yes, sum() can operate on logicals
>> sum([true,true,false])
ans =
2
##### 10 件のコメント8 件の古いコメントを表示8 件の古いコメントを非表示
Mirthand 2021 年 4 月 5 日
You are right, I need to think carefully about how I want to change column 2 so that it works.
Mirthand 2021 年 4 月 5 日

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### その他の回答 (1 件)

Steven Lord 2021 年 4 月 2 日
Use discretize or findgroups to identify groups based on the values in column 2, then use groupsummary or splitapply to perform group processing on column 4 based on the groups you identified in the first step.
##### 1 件のコメント-1 件の古いコメントを表示-1 件の古いコメントを非表示
Mirthand 2021 年 4 月 2 日
Would discretize work if there are zeros?
for example, I still want to group the third row with the group that is 2.
The 7th row, containing zero would also be grouped with 5.
trials = {[ 0, 2, 0, 1, 0, 0, 0
0, 2, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 5, 0, 1, 0, 0, 0
0, 5, 0, 0, 0, 0, 0
0, 5, 0, 1, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 5, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 6, 0, 1, 0, 0, 0 ]};

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