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问题:
I have a data.frame like this:
df <- read.csv(text = "ONE,TWO,THREE
23,234,324
34,534,12
56,324,124
34,234,124
123,534,654")
I want to produce a percent bar plot which looks like this (made in LibreOffice Calc):
Thus, the bars should be standarized so all stacks have the same height and sums to 100%. So far all I have been able to get is is a stacked barplot (not percent), using:
barplot(as.matrix(df))
Any help?
回答1:
Since you tagged this with ggplot2 here's a solution using that package (version 0.9.0) in addition to what you've gotten so far.
We use the position
argument of geom_bar
set to position = "fill"
. You may also use position = position_fill()
if you want to use the arguments of position_fill()
(vjust
and reverse
).
Note that your data is in a 'wide' format, whereas ggplot2
requires it to be in a 'long' format. Thus, we first need to melt
the data.
dat <- read.table(text = " ONE TWO THREE
1 23 234 324
2 34 534 12
3 56 324 124
4 34 234 124
5 123 534 654",sep = "",header = TRUE)
#Add an id variable for the filled regions
library(reshape)
datm <- melt(cbind(dat, ind = rownames(dat)), id.vars = c('ind'))
library(scales)
ggplot(datm,aes(x = variable, y = value,fill = ind)) +
geom_bar(position = "fill",stat = "identity") +
# or:
# geom_bar(position = position_fill(), stat = "identity")
scale_y_continuous(labels = percent_format())
回答2:
prop.table is a nice friendly way of obtaining proportions of tables.
m <- matrix(1:4,2)
m
[,1] [,2]
[1,] 1 3
[2,] 2 4
Leaving margin blank gives you proportions of the whole table
prop.table(m, margin=NULL)
[,1] [,2]
[1,] 0.1 0.3
[2,] 0.2 0.4
Giving it 1 gives you row proportions
prop.table(m, 1)
[,1] [,2]
[1,] 0.2500000 0.7500000
[2,] 0.3333333 0.6666667
And 2 is column proportions
prop.table(m, 2)
[,1] [,2]
[1,] 0.3333333 0.4285714
[2,] 0.6666667 0.5714286
回答3:
Chris Beeley is rigth, you only need the proportions by column. Using your data is:
your_matrix<-(
rbind(
c(23,234,324),
c(34,534,12),
c(56,324,124),
c(34,234,124),
c(123,534,654)
)
)
barplot(prop.table(your_matrix, 2) )
Gives:
回答4:
You just need to divide each element by the sum of the values in its column.
Doing this should suffice:
data.perc <- apply(data, 2, function(x){x/sum(x)})
Note that the second parameter tells apply
to apply the provided function to columns (using 1 you would apply it to rows). The anonymous function, then, gets passed each data column, one at a time.