CODEWITHSUNDEEP
rapplylapply
Moritz
Moritz

Reputation: 309

using sweep (or something similar) in R for listed data frames

I'm doing a handful of transformation steps for several dfs, so I have ventured into the beautiful world of apply, lapply, sweep, etc. Unfortunately I got stuck trying to use sweep for listed dfs.

What I would like to do, is calculate the percentage of each value, based on the mean of each data frame's first row.

So I put my dfs into a list which ends up looking something like this;

df1 <- read.table(header = TRUE, text = "a        b
1 16.26418 19.60232
2 16.09745 18.44320
3 17.25242 18.21141
4 17.61503 17.64766
5 18.35453 19.52620")

df2 <- read.table(header = TRUE, text = "a        b
1 4.518654 4.346056
2 4.231176 4.175854
3 2.658694 4.999478
4 3.348019 2.345594
5 3.103378 2.556690")

list.one <- list(df1,df2)

> list.one
[[1]]
         a        b
1 16.26418 19.60232
2 16.09745 18.44320
3 17.25242 18.21141
4 17.61503 17.64766
5 18.35453 19.52620

[[2]]
         a        b
1 4.518654 4.346056
2 4.231176 4.175854
3 2.658694 4.999478
4 3.348019 2.345594
5 3.103378 2.556690

Now I calculate the mean of each first row and store it

  one.hundred <- lapply(list.one, function(i) 
    {rowMeans(i[1,], na.rm=T)})

> one.hundred
[[1]]
       1 
17.93325 

[[2]]
       1 
4.432355

Now I calculate their percentage (as compared to the values stored in the second list) and the best I came up with is this rather tedious workaround:

df1.per<-sweep(list.one[[1]], 1, one.hundred[[1]], 
               function(x,y){100/y*x})
df2.per<-sweep(list.one[[2]], 1, one.hundred[[2]], 
               function(x,y){100/y*x})

list.new(df1.per,df2.per)

If somebody could suggest me simpler, preferably list based solution that would be great help.

Thanks a lot.

Upvotes: 0

Views: 325

Answers (1)

talat
talat

Reputation: 70316

Here's another approach with sapply and Map that will also return a list of data.frames:

means <- sapply(list.one, function(df) rowMeans(df[1, ], na.rm = TRUE))
Map(function(vec, df) df/vec*100, means, list.one)
#$`1`
#          a         b
#1  90.69287 109.30713
#2  89.76315 102.84360
#3  96.20353 101.55109
#4  98.22553  98.40748
#5 102.34916 108.88266
#
#$`1`
#          a         b
#1 101.94702  98.05298
#2  95.46113  94.21299
#3  59.98378 112.79507
#4  75.53589  52.91981
#5  70.01646  57.68243

data:

> dput(list.one)
list(structure(list(a = c(16.26418, 16.09745, 17.25242, 17.61503, 
18.35453), b = c(19.60232, 18.4432, 18.21141, 17.64766, 19.5262
)), .Names = c("a", "b"), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5")), structure(list(a = c(4.518654, 4.231176, 
2.658694, 3.348019, 3.103378), b = c(4.346056, 4.175854, 4.999478, 
2.345594, 2.55669)), .Names = c("a", "b"), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5")))

Upvotes: 1

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