フィルターのクリア

Help Required in using the function 'removerows' and its reverse function

2 ビュー (過去 30 日間)
Avijit Paul
Avijit Paul 2023 年 12 月 1 日
コメント済み: Avijit Paul 2023 年 12 月 2 日
Hello
I have column matrix 'rain'
d=find(rain>=1.0); %Index of rainfall more than 1mm/h
[x,position]=removerows(rain,'ind',d);
high_rain=rain(rain>=1.0); %Column matrix with only high rains
high_rain=high_rain*2.5; % Manipulate High_rain
%manipulate_rain=removerows('reverse',x,position);
I wanted to reinsert the high_rain back to rain variable using the reverse function of removerows. How to do that.

採用された回答

Image Analyst
Image Analyst 2023 年 12 月 1 日
You can't. See:
x1 = [1 2 4; 1 1 1; 3 2 2; 0 0 0]
x1 = 4×3
1 2 4 1 1 1 3 2 2 0 0 0
[y1,ps] = removerows(x1,'ind',[2 4])
y1 = 2×3
1 2 4 3 2 2
ps = struct with fields:
name: 'removerows' xrows: 4 yrows: 2 remove_ind: [2 4] keep_ind: [1 3] no_change: 0
% Reverse the processing of y1 to get x1 again.
x1_again = removerows('reverse',y1,ps)
x1_again = 4×3
1 2 4 NaN NaN NaN 3 2 2 NaN NaN NaN
Even the help says:
In the reverse calculation, the unknown values of replaced rows are represented with NaN values.
If you want rain, then just keep it in scope. If, for some reason, you want to multiply high rain values by some factor, you can use masking:
rain = 5 * rand(10, 1)
mask = (rain>=1.0) %Index of rainfall more than 1mm/h
rain(mask) = rain(mask) * 2.5

その他の回答 (0 件)

カテゴリ

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

Community Treasure Hunt

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

Start Hunting!

Translated by