将R的“输出”转换为“命令到数据帧 [英] converting output of R's "by" command to data frame
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问题描述
我试图通过命令使用R的来获取数据帧子集的列。例如,考虑这个数据框架:
> z = data.frame(labels = c(a,a,b,c,c),data = matrix(1:20,nrow = 5))
> z
标签数据1 data.2 data.3 data.4
1 a 1 6 11 16
2 a 2 7 12 17
3 b 3 8 13 18
4 c 4 9 14 19
5 c 5 10 15 20
我可以使用R的通过
命令获取列表示根据标签列:
> by(z [,2:5],z $ labels,colMeans)
z [,1]:a
data.1 data.2 data.3 data.4
1.5 6.5 11.5 16.5
---------------------------------------------- --------------
z [,1]:b
data.1 data.2 data.3 data.4
3 8 13 18
------------------------------------------------ ------------
z [,1]:c
data.1 data.2 data.3 data.4
4.5 9.5 14.5 19.5
但是如何强制输出回到数据帧? as.data.frame
不工作...
> ; as.data.frame(by(z [,2:5],z $ labels,colMeans))
as.data.frame.default中的错误(by(z [,2:5],z $ labels ,colmeans)):
不能强制类'by'into a data.frame
解决方案
您可以使用 ddply
从 plyr
包
library(plyr)
pre>
ddply(z,。(labels),numcolwise(mean))
labels data.1 data。 2 data.3 data.4
1 a 1.5 6.5 11.5 16.5
2 b 3.0 8.0 13.0 18.0
3 c 4.5 9.5 14.5 19.5
或
聚合
从stats
aggregate(z [, - 1],by = list(z $ labels),mean)
Group.1 data.1 data.2 data.3 data.4
1 a 1.5 6.5 11.5 16.5
2 b 3.0 8.0 13.0 18.0
3 c 4.5 9.5 14.5 19.5
或
dcast
从reshape2
/ p>
l ibrary(reshape2)
dcast(melt(z),labels〜variable,mean)
使用
sapply
:t(sapply(split(z [, - 1],z $ labels),colMeans))
data.1 data.2 data.3 data.4
a 1.5 6.5 11.5 16.5
b 3.0 8.0 13.0 18.0
c 4.5 9.5 14.5 19.5
I'm trying to use R's
by
command to get column means for subsets of a data frame. For example, consider this data frame:> z = data.frame(labels=c("a","a","b","c","c"),data=matrix(1:20,nrow=5)) > z labels data.1 data.2 data.3 data.4 1 a 1 6 11 16 2 a 2 7 12 17 3 b 3 8 13 18 4 c 4 9 14 19 5 c 5 10 15 20
I can use R's
by
command to get the column means according to the labels column:> by(z[,2:5],z$labels,colMeans) z[, 1]: a data.1 data.2 data.3 data.4 1.5 6.5 11.5 16.5 ------------------------------------------------------------ z[, 1]: b data.1 data.2 data.3 data.4 3 8 13 18 ------------------------------------------------------------ z[, 1]: c data.1 data.2 data.3 data.4 4.5 9.5 14.5 19.5
But how do I coerce the output back to a data frame?
as.data.frame
doesn't work...> as.data.frame(by(z[,2:5],z$labels,colMeans)) Error in as.data.frame.default(by(z[, 2:5], z$labels, colMeans)) : cannot coerce class '"by"' into a data.frame
解决方案You can use
ddply
fromplyr
packagelibrary(plyr) ddply(z, .(labels), numcolwise(mean)) labels data.1 data.2 data.3 data.4 1 a 1.5 6.5 11.5 16.5 2 b 3.0 8.0 13.0 18.0 3 c 4.5 9.5 14.5 19.5
Or
aggregate
fromstats
aggregate(z[,-1], by=list(z$labels), mean) Group.1 data.1 data.2 data.3 data.4 1 a 1.5 6.5 11.5 16.5 2 b 3.0 8.0 13.0 18.0 3 c 4.5 9.5 14.5 19.5
Or
dcast
fromreshape2
packagelibrary(reshape2) dcast( melt(z), labels ~ variable, mean)
Using
sapply
:t(sapply(split(z[,-1], z$labels), colMeans)) data.1 data.2 data.3 data.4 a 1.5 6.5 11.5 16.5 b 3.0 8.0 13.0 18.0 c 4.5 9.5 14.5 19.5
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