[英]Add a different number of columns to dataframe depending on value from row in R
您能帮我为数据框的行添加不同数量的列吗?
例如,我有此DataFrame:
employee <- c('John','Peter','Jolie', 'Katy', 'Pauline')
numberofmonths <- c(7, 5, 11, 3, 12)
employers <- data.frame(employee,numberofmonths)
employee numberofmonths
1 John 7
2 Peter 5
3 Jolie 11
4 Katy 3
5 Pauline 12
现在,我想向employers
添加“雇员”名称所包含的字符列。
所以我想要这样的事情:
employee numberofmonths i i i i i i i
1 John 7 A1 A2 A3 A4 NA NA NA
2 Peter 5 A1 A2 A3 A4 A5 NA NA
3 Jolie 11 A1 A2 A3 A4 A5 NA NA
4 Katy 3 A1 A2 A3 A4 NA NA NA
5 Pauline 12 A1 A2 A3 A4 A5 A6 A7
我已经尝试过以下脚本:
for (i in (1:nrow(employers))) {
for (j in nchar(as.vector(employers[i,]$employee))){
employers<-cbind(employers, i=paste("A", i, sep=""))
}}
但是,与其为约翰和...提供A1:A4,还不为Pauline提供A1:A7,而是为所有人提供A1:A5:
employee numberofmonths i i i i i
1 John 7 A1 A2 A3 A4 A5
2 Peter 5 A1 A2 A3 A4 A5
3 Jolie 11 A1 A2 A3 A4 A5
4 Katy 3 A1 A2 A3 A4 A5
5 Pauline 12 A1 A2 A3 A4 A5
当然,对于名称字符比其他字符少的名称,我们将使用NA
列。 我正在处理带有许多字符串的大数据框,因此任何手动操作都可以使用。 这仅是示例,因此Ai值没有任何意义。
这是使用plyr的解决方案:
require(plyr)
cbind(employers, rbind.fill.matrix(lapply(nchar(employee), function(z) t(paste0("A", 1:z)))))
尝试这个:
nc<-nchar(as.character(employers$employee))
mat<-matrix(NA_character_,nrow=nrow(employers),ncol=max(nc))
indices<-sequence(nc)
values<-paste0("A",indices)
mat[cbind(rep(1:nrow(employers),nc),indices)]<-values
cbind(employers,mat)
# employee numberofmonths 1 2 3 4 5 6 7
#1 John 7 A1 A2 A3 A4 <NA> <NA> <NA>
#2 Peter 5 A1 A2 A3 A4 A5 <NA> <NA>
#3 Jolie 11 A1 A2 A3 A4 A5 <NA> <NA>
#4 Katy 3 A1 A2 A3 A4 <NA> <NA> <NA>
#5 Pauline 12 A1 A2 A3 A4 A5 A6 A7
employee <- c('John','Peter','Jolie', 'Katy', 'Pauline')
numberofmonths <- c(7, 5, 11, 3, 12)
employers <- data.frame(employee,numberofmonths)
employers$employee <- as.character(employers$employee)
emp_app <- as.data.frame(matrix(NA,
nrow = nrow(employers),
ncol = max(nchar(employers$employee))))
for (i in seq_len(nrow(employers))) {
nm_lngth <- nchar(employers$employee)[i]
nm_string <- paste0("A", seq_len(nm_lngth))
for (j in seq_len(nm_lngth)) {
emp_app[i, j] <- nm_string[j]
}
}
employers <- cbind(employers, emp_app)
可能不是最好的解决方案,但它可行
employee <- c('John','Peter','Jolie', 'Katy', 'Pauline')
numberofmonths <- c(7, 5, 11, 3, 12)
employers <- data.frame(employee,numberofmonths)
max = max(nchar(as.character(employers[,1])))
for (c in 1:max) {
employers[,c+2] = ifelse(nchar(as.character(employers[,1]))>=c, paste0("A",c), NA)
}
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