rounding a column vector

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Ankit Labh
Ankit Labh 2018 年 8 月 14 日
コメント済み: Brent F 2021 年 7 月 23 日
I have a column vector with 8000 entries around temp (see array below).
temp=[5 50 150 180 200 220 240 250 260 265 270 272 274 275 275.5 276 277 278 279 279.5 280 280.5 281 281.5 282 284 286 288 290 295 300 310 320]; %temperatures [K]
I did the rounding up to one decimal places. Now I have numbers like 273.9 and 274.1 along with 274. I want to make all such 273.9 and 274.1 to 274, and wherever there is any number with .1 and .9, it should do the floor and ceil with that so that I always get an integer. 275.5 type numbers are the exception.
I tried to use this :
for i= 2:8351
while data(i,2)~=ceil(data(i,2))
if data(i+1,2)== (data(i,2))+0.1
data(i,2)=floor(data(i,2),0);
elseif data(i-1,2)== (data(i,2))-0.1
data(i,2)=ceil(data(i,2),0);
elseif data(i+1,2)== (data(i,2))-0.1
data(i,2)=ceil(data(i,2),0);
elseif data(i-1,2)== (data(i,2))+0.1
data(i,2)=floor(data(i,2),0);
else
data(i,2)=round(data(i,2),1);
end
end
end
Summary: There are 32 temperatures and around 8000 data points near those 32 temperatures. I want to go to one by one in each column and see if it has .1 or .9 and if so, round(or ceil or floor) it to the nearest integer.
Please help me in programming this.
Thanks.
  1 件のコメント
Ankit Labh
Ankit Labh 2018 年 8 月 14 日
Question is not specific to 274 but for all temp.

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採用された回答

Stephen23
Stephen23 2018 年 8 月 14 日
編集済み: Stephen23 2018 年 8 月 14 日
This will round values from (X-1).9 to (X).1 to the integer X, otherwise leave the rounded values alone:
>> vec = 1:0.1:2
vec =
1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00
>> new = round(vec*10)/10;
>> tmp = round(vec*4);
>> idx = mod(tmp,4)==0;
>> new(idx) = tmp(idx)/4
new =
1.00 1.00 1.20 1.30 1.40 1.50 1.60 1.70 1.80 2.00 2.00
  4 件のコメント
Brent F
Brent F 2021 年 7 月 23 日
Clever. @Stephen Cobeldick used a multiplier + round (which rounds in both directions) to leave the middle untouched! I sense a general principle here...

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その他の回答 (1 件)

M
M 2018 年 8 月 14 日
編集済み: M 2018 年 8 月 14 日
Try this:
temp=[273.9 274.1 274.5];
round(temp*2)/2
ans =
274.0000 274.0000 274.5000
  1 件のコメント
Ankit Labh
Ankit Labh 2018 年 8 月 14 日
編集済み: Ankit Labh 2018 年 8 月 14 日
Thank you for your reply. There are 32 temperatures and around 8000 data points near to those 32 temperatures. I want to go to one by one in the column and see if it has .1 or .9 and if so, round(or ceil or floor) it to the nearest integer.

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