Get mean and stdev from all values

7 ビュー (過去 30 日間)
Dion Theunissen
Dion Theunissen 2021 年 6 月 3 日
コメント済み: Steven Lord 2021 年 6 月 3 日
Hi,
I have a list like below. in the first column I have values from 0 to 91 (there can miss values).
0 3
0 4.50000000000000
0 1.37500000000000
0 5
0 3
0 0.838961038961039
0 2.80000000000000
1 1.50000000000000
1 0.555555555555556
1 1.71428571428571
1 3.38666666666667
1 4
1 3.38666666666667
2 1.73684210526316
2 1
2 1.30769230769231
2 2.73333333333333
2 1
2 2
I want to calculate the mean and standard deviation for all of these values if they exist.
How can I realise this? Now I have the folowing:
clear all; clc; close all;
standaard = readmatrix('/Users/diontheunissen/Documents/Apployee/Smart_driver/avgFuel.xlsx');
indices = find(standaard(:,1)==0);
standaard(indices,:) = [];
indices = find(standaard(:,2)>8);
standaard(indices,:) = [];
standaard(:,1) = round(standaard(:,1));
standaard = sortrows(standaard,1);
c = {};
for i = 0:5:85
if i == 0
lower = find(standaard(:,1) == i);
n = i+5;
upper = find(standaard(:,1) == n);
elseif i>1
lower = find(standaard(:,1) == i+1);
n = i+5;
upper = find(standaard(:,1) == n);
end
tab = standaard(lower(1):upper(end),:);
c{i+1} = tab;
end
referent2 = []
for j = 1:5:86
data = cell2mat(c(j));
referent(1,1) = data(1,1);
referent(1,2) = mean(data(:,2));
referent(1,3) = std(data(:,2));
referent2 = [referent2;referent]
end
snelheid = linspace(min(referent2(:,1)),max(referent2(:,1)));
a1 = interp1(referent2(:,1), referent2(:,2:end), snelheid, 'makima');
verbruik = a1(:,1);
upper = a1(:,1)+a1(:,2);
lower = a1(:,1)-a1(:,2);
figure
hold on
plot(snelheid, verbruik, '-b');
plot(snelheid, upper,'-r');
plot(snelheid, lower,'-g');
How can i change this script that i get the mean and stdev from all existing values?
thanks
  2 件のコメント
DGM
DGM 2021 年 6 月 3 日
Can you clarify the structure of the file and what you're trying to do? I'm assuming you're trying to find the blockwise mean and std for column 2, the blocks being described by column 1. Are all blocks the same size? Are the values in column 1 sorted?
Dion Theunissen
Dion Theunissen 2021 年 6 月 3 日
Indeed, the blocks have not the same size and are sorted in column 1.
So i need to check for each value from 1 untill 91 the mean and standard deviation

サインインしてコメントする。

採用された回答

DGM
DGM 2021 年 6 月 3 日
編集済み: DGM 2021 年 6 月 3 日
If column 1 is sorted:
% build test array
idx = repelem(1:10,randi(6,10,1));
idx = idx(1:20);
D = [idx.' rand(20,1)]; % the test array
% find length of blocks
blocklengths = diff([0 find(diff(idx)) numel(idx)]);
% split and process
C = mat2cell(D(:,2),blocklengths,1);
bkmean = cellfun(@mean,C)
bkmean = 6×1
0.8030 0.6236 0.4651 0.2942 0.4628 0.6299
bkstd = cellfun(@std,C)
bkstd = 6×1
0.1993 0.3422 0.1443 0.2592 0.5516 0.0849

その他の回答 (1 件)

Stephan
Stephan 2021 年 6 月 3 日
編集済み: Stephan 2021 年 6 月 3 日
Use a table and then use findgroups combined with splitappy. This will allow you to solve the problem with a few lines of code. This will also work if the columns are not sorted.
  1 件のコメント
Steven Lord
Steven Lord 2021 年 6 月 3 日
findgroups can operate on numeric data as well as tabular data.
But rather than call those two functions sequentially, I'd probably just use groupsummary.
X = [0 3;
0 4.50000000000000;
0 1.37500000000000;
0 5;
0 3;
0 0.838961038961039;
0 2.80000000000000;
1 1.50000000000000;
1 0.555555555555556;
1 1.71428571428571;
1 3.38666666666667;
1 4;
1 3.38666666666667;
2 1.73684210526316;
2 1;
2 1.30769230769231;
2 2.73333333333333;
2 1;
2 2];
[M, BG] = groupsummary(X(:, 2), X(:, 1), @mean)
M = 3×1
2.9306 2.4239 1.6296
BG = 3×1
0 1 2
check = mean(X(X(:, 1)== 1, 2)) % This is the same as M(BG == 1)
check = 2.4239
S = groupsummary(X(:, 2), X(:, 1), @std)
S = 3×1
1.5037 1.3554 0.6728
MS = groupsummary(X(:, 2), X(:, 1), {@mean, @std}) % Compute M and S simultaneously
MS = 3×2
2.9306 1.5037 2.4239 1.3554 1.6296 0.6728

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeNumeric Types についてさらに検索

タグ

製品


リリース

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by