Hi Juvin,
The error "dim(X) must have a positive length" usually shows when you
are passing a vector to "apply", ie.
apply(1:5,2,mean)
#Error in apply(1:5, 2, mean) : dim(X) must have a positive length
Also, if your dataset originally has "1206" columns, it is not
clear why you needed the below code. ("rainfall" is already a
"data.frame")
precip=data.frame(rainfall[1:1206])
Based on the data provided,
rainfall <- read.table(text="1 2 3 4 5 6 7 8
9 10 11
NA 0 0 0 0 12 0 0 0 0 0
NA 0 0 0 0 0 0 0 0 0 0
NA 0 0 0 0 14 0 0 0 0 5
NA 0 0 0 0 0 0 0 0 0 0
NA 0 0 27 0 0 0 0 20 0 165
NA 0 88 38 0 0 0 0 0 0 26
NA 12 12 0 0 0 0 0 0 0 2
NA 2 0 0 0 0 0 0 0 0 0
NA 2 0 0 0 0 0 0 0 0 0
NA 0 24 1 0 0 0 0 3 0 62
NA 26 0 0 0 0 0 0 0 0 33",sep="",
header=TRUE, check.names=FALSE)
apply(rainfall, 2, function(x) c(mean=mean(x, na.rm=TRUE),
median=median(x, na.rm=TRUE), max=max(x, na.rm=TRUE)))
#1 2 3 4 5 6 7 8 9 10 11
#mean NaN 3.818182 11.27273 6 0 2.363636 0 0 2.090909 0 26.63636
#median NA 0.000000 0.00000 0 0 0.000000 0 0 0.000000 0 2.00000
#max -Inf 26.000000 88.00000 38 0 14.000000 0 0 20.000000 0 165.00000
Or using `colMaxs`, `colMedians` from `matrixStats`
library(matrixStats)
rbind(mean=colMeans(rainfall, na.rm=TRUE), median=
colMedians(as.matrix(rainfall),
na.rm=TRUE), max=colMaxs(rainfall, na.rm=TRUE))
Another option would be to use `summarise_each` from `dplyr`
library(dplyr)
rainfall %>%
summarise_each(funs(mean=mean(., na.rm=TRUE), median=median(.,
na.rm=TRUE),
max=max(., na.rm=TRUE)))
A.K.
I tried to calculate a mean from a csv table by forming a data frame,
but it says dim(x)must have a positive length. The table has 1206 column and 31
rows. I want to calculate mean, median, and maximum from the the table. The
table has some NA values which i dont want to include. The
table looks as follows:
1 2 3 4 5 6 7 8 9 10 11
NA 0 0 0 0 12 0 0 0 0 0
NA 0 0 0 0 0 0 0 0 0 0
NA 0 0 0 0 14 0 0 0 0 5
NA 0 0 0 0 0 0 0 0 0 0
NA 0 0 27 0 0 0 0 20 0 165
NA 0 88 38 0 0 0 0 0 0 26
NA 12 12 0 0 0 0 0 0 0 2
NA 2 0 0 0 0 0 0 0 0 0
NA 2 0 0 0 0 0 0 0 0 0
NA 0 24 1 0 0 0 0 3 0 62
NA 26 0 0 0 0 0 0 0 0 33
I used following code to calculate mean:
Any help would be appreciated.
rainfall=read.table('bmark.csv',header=T,sep=',')
precip=data.frame(rainfall[1:1206])
monthlyMean=apply(precip, MARGIN=2,FUN=mean,na.rm=TRUE)
Juvin