转置数据集

Question

我想转置类似于my.data下面的数据集,然后对行进行求和.

my.data <- "landuse units year county.a  county.b  county.c  county.d 
            apple   acres 2010     0         2         4         6 
            pear    acres 2010    10        20        30        40
            peach   acres 2010   500       400       300       200"

my.data2 <- read.table(textConnection(my.data), header = T)
my.data2

所需的输出是:

 counties all.fruit
 county.a       510
 county.b       422
 county.c       334
 county.d       246

我可以使用下面的代码执行此操作.但是,以下代码似乎必须是巨大的矫枉过正.我希望有一个更简单的解决方案.

# transpose the data set

tmy.data2 <- t(my.data2)
tmy.data2 <- as.data.frame(tmy.data2)

# assign row names to the data set

my.rows <- row.names(tmy.data2)

transposed.data <- cbind(my.rows, tmy.data2)
transposed.data

# extract numbers to obtain row sums

fruit.data <- as.data.frame(transposed.data[4:dim(transposed.data)[1], 2:dim(transposed.data)[2]])

fruit.data2 <- as.matrix(fruit.data)

fruit.data3 <- matrix(as.numeric(fruit.data2), nrow=( dim(fruit.data2)[1] ),  byrow=F)

# sum fruit by county

all.fruit <- rowSums(fruit.data3, na.rm=T)

# create row names for summed fruit data

counties <- my.rows[4:length(my.rows)]

almost.final.data <- cbind(counties, all.fruit)

really.final.data <- as.data.frame(almost.final.data)

really.final.data[,2] <- as.numeric(as.character(really.final.data[,2]))
really.final.data

str(really.final.data)

谢谢你的任何建议.我可以使用上面的代码,但将此请求视为一个大大改进我的编程的机会.

Answer 1

你为什么不能只添加列？

colSums(my.data2[, 4:7]) 

要么

library(plyr)
numcolwise(sum)(my.data2)
  year county.a county.b county.c county.d
1 6030      510      422      334      246
> 

也就是说,如果你想重新组织,有很多选择.该reshape2包提供了令人愉快的语法

library(reshape2)
> my.data.melt <- melt(my.data2, id.vars=c('units', 'year', 'landuse'))
> my.data.melt
   units year landuse variable value
1  acres 2010   apple county.a     0
2  acres 2010    pear county.a    10
3  acres 2010   peach county.a   500
4  acres 2010   apple county.b     2
5  acres 2010    pear county.b    20
6  acres 2010   peach county.b   400
7  acres 2010   apple county.c     4
8  acres 2010    pear county.c    30
9  acres 2010   peach county.c   300
10 acres 2010   apple county.d     6
11 acres 2010    pear county.d    40
12 acres 2010   peach county.d   200

然后我会用plyr:

> library(plyr)
> ddply(my.data.melt, .(variable), summarise, all.fruit=sum(value))
  variable all.fruit
1 county.a       510
2 county.b       422
3 county.c       334
4 county.d       246
> 

您也可以使用base R aggregate或data.table包来完成此操作.

data.table

> library(data.table)
> my.data.melt <- as.data.table(melt(my.data2, id.vars=c('units', 'year', 'landuse')))
> my.data.melt[,list(all.fruit = sum(value)), by = variable]
   variable all.fruit
1: county.a       510
2: county.b       422
3: county.c       334
4: county.d       246

或者,如果你希望它留在宽格式

> DT <- as.data.table(my.data2)
> DT[, lapply(.SD, sum, na.rm=TRUE), .SDcols = grep("county",names(DT))])
   county.a county.b county.c county.d
1:      510      422      334      246

# NB: This needs v1.8.3. Before that, an as.data.table() call was required as
# the lapply(.SD,...) used to return a named list in this no grouping case.

骨料

> aggregate(value~variable, my.data.melt, sum)
  variable value
1 county.a   510
2 county.b   422
3 county.c   334
4 county.d   246

Answer 2

我只是对"county"列进行子集,求和,并使用结果创建data.frame:

out <- colSums(my.data2[,grepl("county",colnames(my.data2))])
out2 <- data.frame(counties=names(out), all.fruit=out,
          row.names=NULL, stringsAsFactors=FALSE)