Help normalizing data in table

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Nina Perf
Nina Perf 2021 年 12 月 3 日
編集済み: Nina Perf 2021 年 12 月 3 日
Hi,
I have a 100x6 Table with 6 variables. I want to center the data so that it has mean 0 and standard deviation 1, using the zscore.
I tried using this in the 3 variables of interest but is not working.
Normalized = normalize(data(:,1:3)));
If I try to convert the table to array then I loose the table information.
Thank you in advance!

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Dave B
Dave B 2021 年 12 月 3 日
編集済み: Dave B 2021 年 12 月 3 日
When you want to work with multiple variables in a table at once, you have to use { } (this is because tables work with heterogenous data)
data=table(10*randn(100,1)+5, randn(100,1)/10, randn(100,1).^2);
ndata = data(:,1:3);
ndata{:,1:3} = normalize(ndata{:,1:3});
round(mean(table2array(ndata)),-10)
ans = 1×3
0 0 0
std(table2array(ndata))
ans = 1×3
1.0000 1.0000 1.0000
  3 件のコメント
Dave B
Dave B 2021 年 12 月 3 日
For the round bit - that's just because (after normalizing) the mean is close to zero, but not exactly zero. I rounded it to 10 digits. Here's what it looks like without the round:
data=table(10*randn(100,1)+5, randn(100,1)/10, randn(100,1).^2);
ndata = data(:,1:3);
ndata{:,1:3} = normalize(ndata{:,1:3});
mean(table2array(ndata))
ans = 1×3
1.0e+-15 * -0.0844 0.0266 -0.1821
So that's like -0.0000000000000000844 for the first one (I might have miscounted my zeros!)
I'm not sure what you mean by the std is not 1, can you give an example? (similar to the means, they're not exactly 1, but I'd think would be close to 1). Note that you can also use zscore in place of normalize, I just wanted to help with the 'how to apply it to table' bit...but I'd expect zscore and normalize to produce the same results...
Nina Perf
Nina Perf 2021 年 12 月 3 日
編集済み: Nina Perf 2021 年 12 月 3 日
Thank you! The zscore function works well. However, the normalize function gives std close to 0..

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