CODEWITHSUNDEEP
rdata.table
rnso
rnso

Reputation: 24535

Calculate summary statistics (e.g. mean) on all numeric columns using data.table

I have data with both numeric and non-numeric columns like this:

mydt
          vnum1 vint1 vfac1 vch1
 1: -0.30159484     8     3    E
 2: -0.09833430     8     1    D
 3: -2.15963282     1     3    D
 4:  0.03904374     5     2    B
 5:  1.54928970     4     1    C
 6: -0.73873654     5     1    A
 7: -0.68594479     9     2    B
 8:  1.35765612     1     2    E
 9:  1.46958351     2     1    B
10: -0.89623979     2     4    E

How can I select only numeric columns and calculate their mean using data.table?

I have tried this: 

mydt[ , lapply(.SD, mean), ]
#        vnum1 vint1 vfac1 vch1
# 1: -0.046491   4.5    NA   NA
# Warning messages:
# 1: In mean.default(X[[3L]], ...) :
#  argument is not numeric or logical: returning NA
# 2: In mean.default(X[[4L]], ...) :
#  argument is not numeric or logical: returning NA


dput(mydt)
structure(list(vnum1 = c(-0.301594844692861, -0.0983343040483769, 
-2.15963282153076, 0.03904374068617, 1.54928969700272, -0.738736535236348, 
-0.685944791146016, 1.35765612481877, 1.46958350568506, -0.896239790653183
), vint1 = c(8L, 8L, 1L, 5L, 4L, 5L, 9L, 1L, 2L, 2L), vfac1 = structure(c(3L, 
1L, 3L, 2L, 1L, 1L, 2L, 2L, 1L, 4L), .Label = c("1", "2", "3", 
"4"), class = "factor"), vch1 = structure(c(5L, 4L, 4L, 2L, 3L, 
1L, 2L, 5L, 2L, 5L), .Label = c("A", "B", "C", "D", "E"), class = "factor")), .Names = c("vnum1", 
"vint1", "vfac1", "vch1"), class = c("data.table", "data.frame"
), row.names = c(NA, -10L), .internal.selfref = <pointer: 0x991c070>)

I have also tried the following, but it does not work: 

 mydt[ , lapply(.SD, is.numeric),]
 #    vnum1 vint1 vfac1  vch1
 # 1:  TRUE  TRUE FALSE FALSE

mydt[,mydt[,lapply(.SD, is.numeric),]]
#    vnum1 vint1 vfac1  vch1
# 1:  TRUE  TRUE FALSE FALSE

mydt[ , mydt[ , lapply(.SD, is.numeric) , ], with = F]
# Error in Math.data.frame(j) : 
#  non-numeric variable in data frame: vnum1vint1vfac1vch1

mydt[ , c(mydt[ , lapply(.SD, is.numeric)), ], with = F]
# Error: unexpected ')' in "mydt[,c(mydt[,lapply(.SD, is.numeric))"

As suggested by @Arun, I tried following but cannot get a subset: 

 xx <- mydt[ , lapply(.SD, is.numeric), ]
 xx
 #    vnum1 vint1 vfac1  vch1
 # 1:  TRUE  TRUE FALSE FALSE

 mydt[ , lapply(.SD, mean), .SDcols = xx]
 # Error in `[.data.table`(mydt, , lapply(.SD, mean), .SDcols = xx) : 
 # .SDcols should be column numbers or names

As suggested by @David, I tried following but get NULL values for non-numeric columns. I want to get a subset of mydt so that other columns are not even listed. 

mydt[ , lapply(.SD, function(x) if(is.numeric(x)) mean(x))]
#       vnum1 vint1 vfac1 vch1
# 1: -0.046491   4.5  NULL NULL

I am missing data.frame: 

sapply(mydf, is.numeric)
# vnum1 vint1 vfac1  vch1 
#  TRUE  TRUE FALSE FALSE 

mydf[sapply(mydf, is.numeric)]
#         vnum1 vint1
#1  -0.30159484     8
#2  -0.09833430     8
#3  -2.15963282     1
#4   0.03904374     5
#5   1.54928970     4
#6  -0.73873654     5
#7  -0.68594479     9
#8   1.35765612     1
#9   1.46958351     2
#10 -0.89623979     2


sapply(mydf[sapply(mydf, is.numeric)], mean)
#    vnum1     vint1 
#-0.046491  4.500000

OK. Thanks to David's comment, following works: 

mydt[, sapply(mydt, is.numeric), with = FALSE][,sapply(.SD, mean),]
#    vnum1     vint1 
# -0.046491  4.500000 

mydt[, sapply(mydt, is.numeric), with = FALSE]
#          vnum1 vint1
# 1: -0.30159484     8
# 2: -0.09833430     8
# 3: -2.15963282     1
# 4:  0.03904374     5
# ...

Upvotes: 5

Views: 5424

Answers (3)

Seyma Kalay
Seyma Kalay

Reputation: 2859

I got the same problem, also the code below may help too.

data("mtcars")
mtcars$X1 <- factor(mtcars$gear, levels = c(4,3,5)); str(mtcars) #create an non numeric column X1
my.mean <- function(x){ if(is.numeric(x)) c(mean(x), median(x))} 
my.df <- setNames(as.data.frame(unlist(lapply(mtcars, FUN = my.mean))), "values"); my.df
my.df$names <- rep(c("mean","median"), times = length(row.names(my.df))/2); my.df
my.df$variables <-  rownames(my.df); my.df
library(stringr)
my.df$variables <- str_remove(my.df$variables, "[12]"); my.df 

data_wide <- spread(my.df,  names, values)
data_wide

> data_wide
   variables       mean  median
1         am   0.406250   0.000
2       carb   2.812500   2.000
3        cyl   6.187500   6.000
4       disp 230.721875 196.300
5       drat   3.596563   3.695
6       gear   3.687500   4.000
7         hp 146.687500 123.000
8        mpg  20.090625  19.200
9       qsec  17.848750  17.710
10        vs   0.437500   0.000
11        wt   3.217250   3.325

Upvotes: 1

Moein
Moein

Reputation: 121

You can use the below format in just one line without having to use sapply:

mydt[, lapply(.SD, mean), .SDcols = is.numeric]

Also, if you're working with real data, there's a good chance your data has values. Here's how it'd work in case of NA's:

mydt[, lapply(.SD, function(i) mean(i, na.rm = T)), .SDcols = is.numeric]

Upvotes: 1

Arun
Arun

Reputation: 118779

By searching on SO for .SDcols, I landed up on this answer, which I think explains quite nicely how to use it.

cols = sapply(mydt, is.numeric)
cols = names(cols)[cols]
mydt[, lapply(.SD, mean), .SDcols = cols]
#        vnum1 vint1
# 1: -0.046491   4.5

Doing mydt[, sapply(mydt, is.numeric), with = FALSE] (note: the "modern" way to do that is mydt[ , .SD, .SDcols = is.numeric])is not that efficient because it subsets your data.table with those columns and that makes a (deep) copy - more memory used unnecessarily.

And using colMeans coerces the data.table into a matrix, which again is not so memory efficient.

Upvotes: 13

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