2

I am an only very occasional R user and this is the first time I am asking a question regarding R here or anywhere else online, so I apologize beforehand if anything remains unclear.

I have a numeric dataframe with about 100 columns in each of which there is the same number (number 10 in this example) that needs to be multiplied with a value from a numeric vector, which is specific to each column. I am completely stuck and would appreciate any help.

Here is a simplified example:

df

                    V1            V2          V3
1                   0             0           2  
2                   1             0           2  
3                   0             0           1  
4                   0             0           2  
5                   0             0           1  
6                  10             0           1  
7                   0             0           1  
8                   0             0           2  
9                   0            10           2  
10                  0             0           2  
11                 10             0           1  
12                  0             0          10  
13                  1             2           1  
14                  0             0           2  
15                  0             0           0  
16                  0             1           2  
17                  1             0          10  
18                  1             1           1  
19                  0             0           1  
20                  0             0           2

The corresponding vector would look as follows:

V

v1                 v2                 v3  
0.01256117         0.03037231         0.55444079

So, the values "10" of df column V1 would need to be multiplied by value v1 of vector V, the values "10" of df column V2 by the value v2 of vector V, etc.

Any help is very much appreciated!

SabDeM
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Chris
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  • it doesn't sound like an hard thing, but you have to provide a (better) piece of your code (a reproducible example) in order to allow user to help you. A desired output would be a plus. [Here](http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) you can find out how to make a reproducible example. – SabDeM Jul 05 '15 at 01:11
  • So, you only want to change the 10's in each column? – Mark Jul 05 '15 at 01:14
  • exactly, but by multiplying with a number specific to the column. – Chris Jul 05 '15 at 01:25

3 Answers3

3

Here is one simple approach:

# sample data
df<-data.frame(v1=c(1:10,10), v2= c(5:13,10,14), v3=8:18)
vec=c(0.1, 0.2, 0.3) # sample vector to multiply by
df
#   v1 v2 v3
#1   1  5  8
#2   2  6  9
#3   3  7 10
#4   4  8 11
#5   5  9 12
#6   6 10 13
#7   7 11 14
#8   8 12 15
#9   9 13 16
#10 10 10 17
#11 10 14 18

df2 <- t(t(df==10) * vec * t(df))
df[df==10] <- 0
df + df2
#   v1 v2 v3
#1   1  5  8
#2   2  6  9
#3   3  7  3
#4   4  8 11
#5   5  9 12
#6   6  2 13
#7   7 11 14
#8   8 12 15
#9   9 13 16
#10  1  2 17
#11  1 14 18
Jota
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  • Frank, thanks a lot, the "simple solution" works just fine with me! Your answer saved me a lot of extra hours...! – Chris Jul 05 '15 at 10:23
2

Here is a variation

df1 <- (df!=10)*df + ((df==10)*df) * vec1[col(df)]
df1
#          V1        V2       V3
#1  0.0000000 0.0000000 2.000000
#2  1.0000000 0.0000000 2.000000
#3  0.0000000 0.0000000 1.000000
#4  0.0000000 0.0000000 2.000000
#5  0.0000000 0.0000000 1.000000
#6  0.1256117 0.0000000 1.000000
#7  0.0000000 0.0000000 1.000000
#8  0.0000000 0.0000000 2.000000
#9  0.0000000 0.3037231 2.000000
#10 0.0000000 0.0000000 2.000000
#11 0.1256117 0.0000000 1.000000
#12 0.0000000 0.0000000 5.544408
#13 1.0000000 2.0000000 1.000000
#14 0.0000000 0.0000000 2.000000
#15 0.0000000 0.0000000 0.000000
#16 0.0000000 1.0000000 2.000000
#17 1.0000000 0.0000000 5.544408
#18 1.0000000 1.0000000 1.000000
#19 0.0000000 0.0000000 1.000000
#20 0.0000000 0.0000000 2.000000

For big datasets, it may also be better to use lapply/Map etc

f1 <- function(x,y) {i <- x==10
                    x[i] <- x[i]*y 
                     x}
df2 <- data.frame(Map(f1, df, vec1))
df2 
#          V1        V2       V3
#1  0.0000000 0.0000000 2.000000
#2  1.0000000 0.0000000 2.000000
#3  0.0000000 0.0000000 1.000000
#4  0.0000000 0.0000000 2.000000
#5  0.0000000 0.0000000 1.000000
#6  0.1256117 0.0000000 1.000000
#7  0.0000000 0.0000000 1.000000
#8  0.0000000 0.0000000 2.000000
#9  0.0000000 0.3037231 2.000000
#10 0.0000000 0.0000000 2.000000
#11 0.1256117 0.0000000 1.000000
#12 0.0000000 0.0000000 5.544408
#13 1.0000000 2.0000000 1.000000
#14 0.0000000 0.0000000 2.000000
#15 0.0000000 0.0000000 0.000000
#16 0.0000000 1.0000000 2.000000
#17 1.0000000 0.0000000 5.544408
#18 1.0000000 1.0000000 1.000000
#19 0.0000000 0.0000000 1.000000
#20 0.0000000 0.0000000 2.000000

identical(df1, df2)
#[1] TRUE

Or an option with data.table

library(data.table)#v1.9.5+
setDT(df)
for(j in seq_along(df)){
 set(df, i=NULL, j=j, value= as.numeric(df[[j]]))
 set(df, i=which(df[[j]]==10), j=j, value= df[[j]][df[[j]]==10]*vec1[j])
}

df
#          V1        V2       V3
#1: 0.0000000 0.0000000 2.000000
#2: 1.0000000 0.0000000 2.000000
#3: 0.0000000 0.0000000 1.000000
#4: 0.0000000 0.0000000 2.000000
#5: 0.0000000 0.0000000 1.000000
#6: 0.1256117 0.0000000 1.000000
#7: 0.0000000 0.0000000 1.000000
#8: 0.0000000 0.0000000 2.000000
#9: 0.0000000 0.3037231 2.000000
#10:0.0000000 0.0000000 2.000000
#11:0.1256117 0.0000000 1.000000
#12:0.0000000 0.0000000 5.544408
#13:1.0000000 2.0000000 1.000000
#14:0.0000000 0.0000000 2.000000
#15:0.0000000 0.0000000 0.000000
#16:0.0000000 1.0000000 2.000000
#17:1.0000000 0.0000000 5.544408
#18:1.0000000 1.0000000 1.000000
#19:0.0000000 0.0000000 1.000000
#20:0.0000000 0.0000000 2.000000

data

df <- structure(list(V1 = c(0L, 1L, 0L, 0L, 0L, 10L, 0L, 0L, 0L, 0L, 
10L, 0L, 1L, 0L, 0L, 0L, 1L, 1L, 0L, 0L), V2 = c(0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 10L, 0L, 0L, 0L, 2L, 0L, 0L, 1L, 0L, 1L, 
0L, 0L), V3 = c(2L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 10L, 
1L, 2L, 0L, 2L, 10L, 1L, 1L, 2L)), .Names = c("V1", "V2", "V3"
), class = "data.frame", row.names = c(NA, -20L))

vec1 <-  c(v1=0.01256117, v2 =0.03037231,v3  =0.55444079)
smci
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akrun
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1

Here's another suggestion:

arr <- which(df==10, arr.ind=TRUE)
df[arr] <- df[arr] * v[arr[,2]]
#> df
#          V1        V2       V3
#1  0.0000000 0.0000000 2.000000
#2  1.0000000 0.0000000 2.000000
#3  0.0000000 0.0000000 1.000000
#4  0.0000000 0.0000000 2.000000
#5  0.0000000 0.0000000 1.000000
#6  0.1256117 0.0000000 1.000000
#7  0.0000000 0.0000000 1.000000
#8  0.0000000 0.0000000 2.000000
#9  0.0000000 0.3037231 2.000000
#10 0.0000000 0.0000000 2.000000
#11 0.1256117 0.0000000 1.000000
#12 0.0000000 0.0000000 5.544408
#13 1.0000000 2.0000000 1.000000
#14 0.0000000 0.0000000 2.000000
#15 0.0000000 0.0000000 0.000000
#16 0.0000000 1.0000000 2.000000
#17 1.0000000 0.0000000 5.544408
#18 1.0000000 1.0000000 1.000000
#19 0.0000000 0.0000000 1.000000
#20 0.0000000 0.0000000 2.000000

data

df <- structure(list(V1 = c(0L, 1L, 0L, 0L, 0L, 10L, 0L, 0L, 0L, 0L,10L, 
0L, 1L, 0L, 0L, 0L, 1L, 1L, 0L, 0L), V2 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 10L, 0L, 0L, 0L, 2L, 0L, 0L, 1L, 0L, 1L, 0L, 0L), 
V3 = c(2L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 10L, 1L, 2L, 0L, 2L,
10L, 1L, 1L, 2L)), .Names = c("V1", "V2", "V3"), class = "data.frame", 
row.names = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", 
"11", "12", "13", "14", "15", "16", "17", "18", "19", "20"))

v <- c(0.01256117, 0.03037231, 0.55444079)
RHertel
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