Match two variables list

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DavidL88
DavidL88 2022-8-19
编辑： Les Beckham 2022-8-19
Is there a way to transalte across these variable lists
c = ["80ms,110ms", "110ms,140ms", "140ms,195ms", "195ms,250ms", "250ms,400ms", "400ms,550ms", "550ms,725ms", "725ms,900ms"]
Times = {[0.08, 0.11], [0.11, 0.14], [0.14, 0.195], [0.195, 0.25], [0.25, 0.4], [0.4, 0.55], [0.55, 0.725], [0.725, 0.9]}'
such that if I have str.Var1 = {'80ms,110ms'} I can convert it to Times(1) = [0.08, 0.11]?
I am using the first variable to inform what the input should be (second variable) in a statistics test. So whatever value str.Var1 matches in c I get the correspomding value in Times.
> str.Var1
ans =
1×1 cell array
{'80ms,110ms'}
>> Times(1)
ans =
1×1 cell array
{1×2 double}

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Voss
Voss 2022-8-19
编辑：Voss 2022-8-19
Ran in:
c = ["80ms,110ms", "110ms,140ms", "140ms,195ms", "195ms,250ms", "250ms,400ms", "400ms,550ms", "550ms,725ms", "725ms,900ms"]
c = 1×8 string array
"80ms,110ms" "110ms,140ms" "140ms,195ms" "195ms,250ms" "250ms,400ms" "400ms,550ms" "550ms,725ms" "725ms,900ms"
c_temp = c + ",";
Times = num2cell(reshape(sscanf([c_temp{:}],'%fms,')/1000,2,[]).',2)
Times = 8×1 cell array
{[0.0800 0.1100]} {[0.1100 0.1400]} {[0.1400 0.1950]} {[0.1950 0.2500]} {[0.2500 0.4000]} {[0.4000 0.5500]} {[0.5500 0.7250]} {[0.7250 0.9000]}

更多回答（1 个）

Les Beckham
Les Beckham 2022-8-19
编辑：Les Beckham 2022-8-19
Ran in:
Perhaps what you want is this:
c = ["80ms,110ms", "110ms,140ms", "140ms,195ms", "195ms,250ms", "250ms,400ms", "400ms,550ms", "550ms,725ms", "725ms,900ms"];
Times = {[0.08, 0.11], [0.11, 0.14], [0.14, 0.195], [0.195, 0.25], [0.25, 0.4], [0.4, 0.55], [0.55, 0.725], [0.725, 0.9]}';
str.Var1 = {'80ms,110ms'};
cellresult = Times(strcmp(str.Var1, c));
result = cellresult{1} % extract the double vector from the cell
result = 1×2
0.0800 0.1100

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