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I'd like to insert a last row to a dataframe that would show either the sum of the positive values or the sum of the negative value per column, whichever sum has the greatest amplitude.

The final table would look like this with the last row ("MaxAmplitude") being the result I would like to obtain:

  c("A", "B", "C", "Max Amplitude")  X1  X2  X3  X4  X5
1                                 A -30  10  10  -2   8
2                                 B   5  50 100 -10 -10
3                                 C -10 -30   1  -1   3
4                     Max Amplitude -40  60 111 -13  11

My idea was a multiple step approach, where I would successively add several rows:

  • 1st row making the sum of the positive values
  • 2nd row making the sum of the negative values
  • 3rd row doing the test of whichever of the row 1 or 2 has the greatest amplitude by comparing the absolute value of both sums.

My issue is that I'd like to do it without having to add that many rows to the dataframe. Besides, I don't know how I could do to diplay the negative sign in case the sum of negative values has the greatest amplitude (I would lose this information if I compare absolute value).

Would someone have any solution as to how I could code this please? Thanks a lot for your help!

df <- cbind(c("A","B","C"),data.frame(matrix(c(-30,5,-10,
                                                       10,50,-40,
                                                       10,100,1,
                                                       -2,-10,-1,
                                                       8,-10,3
                                                       ), nrow = 3, ncol = 5)))
h3rm4n
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user8453031
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1 Answers1

2

Is this what you want ? Also , apply return a matrix not data.frame

apply(df[,2:6],2,function(x) sum(x[which((x>0)==(sum(x)>0))]))


 X1  X2  X3  X4  X5 
-40  60 111 -13  11

Just df[nrow(df)+1,]=c('Max Amplitude',c(apply(df[,2:6],2,function(x) sum(x[which((x>0)==(sum(x)>0))]))))

> df
  c("A", "B", "C")  X1  X2  X3  X4  X5
1                A -30  10  10  -2   8
2                B   5  50 100 -10 -10
3                C -10 -40   1  -1   3
4    Max Amplitude -40  60 111 -13  11
BENY
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  • Many thanks Wen, this is exactly what I wanted although I must admit I didn't fully understand how the code got the correct result. In particular, I'm not sure to understand the logic behind: sum(x[which((x>0)==(sum(x)>0))]). – user8453031 Aug 22 '17 at 00:35
  • @Had57 is sum(x)>0 which means positive is large than negative , then base on your explanation you should chose positive number and sum them together, is that clear? – BENY Aug 22 '17 at 00:36