Question

我们说我有以下data.frame，它将R包的名称与它所属的CRAN任务视图相关联：

dictionary <- data.frame(task.view = c(rep("High.Performance.Computing", 3), rep("Machine.Learning", 3)), package = c("Rcpp", "HadoopStreaming", "rJava", "e1071", "nnet", "RWeka"))

#                   task.view         package
#  High.Performance.Computing            Rcpp
#  High.Performance.Computing HadoopStreaming
#  High.Performance.Computing           rJava
#            Machine.Learning           e1071
#            Machine.Learning            nnet
#            Machine.Learning           RWeka

然后我计算从学生写的四个工具之一调用每个包的次数：

package.referals <- data.frame(Rcpp = c(1, 0, 1, 1), HadoopStreaming = c(1, 0, 0, 0),  rJava = c(1, 0, 0, 1), e1071 = c(1, 1, 1, 1), nnet = c(1, 0, 0, 0), RWeka = c(1, 0, 0, 1), row.names = paste("student pkg", 1:4))

#               Rcpp HadoopStreaming rJava e1071 nnet RWeka
# student pkg 1    1               1     1     1    1     1
# student pkg 2    0               0     0     1    0     0
# student pkg 3    1               0     0     1    0     0
# student pkg 4    1               0     1     1    0     1

如何根据包任务视图关系的data.frame重构上面的package.referals data.frame的列？

E.g。我想输出

data.frame(High.Performance.Computing = c(3, 0, 1, 2), Machine.Learning = c(3, 1, 1, 2), row.names = paste("student pkg", 1:4))

#               High.Performance.Computing Machine.Learning
# student pkg 1                          3                3
# student pkg 2                          0                1
# student pkg 3                          1                1
# student pkg 4                          2                2

我尝试了以下但是在尝试将其重组为我想要的输出（求和和转置）时我遇到了困难：

require(data.table)

# column names of package.referals data.frame
package.referals.colnames <- names(package.referals)

# a data.table of my task view and package relations, keyed by package name
dictionary.dt <- data.table(dictionary, key = "package")

# a data.table of my package.referals data.frame, transposed, and keyed by package name
package.referals.dt <- data.table(package = package.referals.colnames, t(package.referals), key="package")

# Joining data.tables so that the package name and corresponding task view are on the same line
dt <- package.referals.dt[J(dictionary.dt)]
setkey(dt, "task.view")

#            package student pkg 1 student pkg 2 student pkg 3 student pkg 4                  task.view
# 1: HadoopStreaming             1             0             0             0 High.Performance.Computing
# 2:            Rcpp             1             0             1             1 High.Performance.Computing
# 3:           rJava             1             0             0             1 High.Performance.Computing
# 4:           e1071             1             1             1             1           Machine.Learning
# 5:            nnet             1             0             0             0           Machine.Learning
# 6:           RWeka             1             0             0             1           Machine.Learning

Answer 1

以下是reshape和基础R：

的解决方案

package.referals$id <- rownames(package.referals)
pkgr <- melt(package.referals, variable.name="package")
pkgr <- pkgr[pkgr$value>0,]
df <- merge(pkgr, dictionary, all.x=TRUE)
table(df$id, df$task.view)

如果您真的想使用data.table代替merge，可以用以下代码替换最后的第三行：

pkgr <- data.table(pkgr, key="package")
dictionary <- data.table(dictionary, key="package")
df <- pkgr[dictionary]
table(df$id, df$task.view)

Answer 2

您可以匹配并重命名package.referals的列，然后在名称相同的列上执行rowSums ...

names( package.referals ) <- dictionary$task.view[ match( names( package.referals ) , dictionary$package ) ]

sapply( unique( names( package.referals ) ) , function(x) rowSums( package.referals[ , names( package.referals ) %in% x ] ) )
#              High.Performance.Computing Machine.Learning
#student pkg 1                          3                3
#student pkg 2                          0                1
#student pkg 3                          1                1
#student pkg 4                          2                2

Answer 3

您还可以在一个data.frame中插入所有信息，然后aggregate：

    dictionary <- data.frame(task.view = c(rep("High.Performance.Computing", 3), rep("Machine.Learning", 3)), package = c("Rcpp", "HadoopStreaming", "rJava", "e1071", "nnet", "RWeka"))

    package.referals <- data.frame(Rcpp = c(1, 0, 1, 1), HadoopStreaming = c(1, 0, 0, 0),  rJava = c(1, 0, 0, 1), e1071 = c(1, 1, 1, 1), nnet = c(1, 0, 0, 0), RWeka = c(1, 0, 0, 1), row.names = paste("student pkg", 1:4))

    pack.ref <- as.data.frame(t(package.referals)) #transpose for easier manipulation

    pack.ref$task.view <- as.character(dictionary$task.view[unlist(lapply(colnames(package.referals), grep, dictionary$package))]) #add column with "task.view" of each package (here is obvious)

    DF <- as.data.frame(t(aggregate(pack.ref[,1:4], by = list(pack.ref$task.view), sum))) #"aggregate"

    DF
    #                                      V1               V2
    #Group.1       High.Performance.Computing Machine.Learning
    #student pkg 1                          3                3
    #student pkg 2                          0                1
    #student pkg 3                          1                1
    #student pkg 4                          2                2

根据名称类型将data.frame的列汇总在一起

3 个答案: