R:根据另一列的值计算一列的值数

I have a bit of an unclear question, so I hope I can explain this properly. I am using R. I know for loops can be slow in R, but for me it would be ok to use a for loop in this case.

我有一个不太清楚的问题，希望我能解释清楚。我使用R，我知道在R中for循环是很慢的，但是对于我来说，在这个例子中使用for循环是可以的。

I have a dataframe like this:

我有一个这样的dataframe:

    id_A    id_B    id_C    calc_A  calc_B  calc_C  
1   x,z     d       g,f        1        1       5
2   x,y,z   d,e     f          1        2       8
3   y,z     d,e     g          6        7       1

I also have a vector with the names c('A', 'B', 'C', etc.) What I want to do is to count for every row, how many id’s have a calc <= 2. id_A is linked to calc_A, etc.

我还有一个名为c的向量(' a '， 'B'， ' c '等等)我要做的是对每一行进行计数，有多少id有一个calc <= 2。id_A链接到calc_A，等等。

For example, for the first row A and B have calc values <= 2, together A and B have 3 id's. So the output will be something like this:

例如，对于第一行A和B有calc值<= 2,A和B有3个id。输出是这样的

   count
1   3
2   5
3   1

3 个解决方案

#1

It's a bit messy, but this should do the trick (for data.frame d):

这有点乱，但这应该可以达到目的(对于d帧来说):

# store indices of calc columns and id columns
calc.cols <- grep('^calc', names(d))
id.cols <- grep('^id', names(d))

sapply(split(d, seq_len(nrow(d))), function(x) {
  length(unique(unlist(strsplit(paste(x[, id.cols][which(x[, calc.cols] <= 2)], 
                                      collapse=','), ','))))
})

# 1 2 3 
# 3 5 1

#2

Assuming that the ID columns and the calc columns are in the same order

假设ID列和calc列的顺序相同

 library(stringr)
 indx <- sapply(df[,1:3], str_count, ",")+1
 indx[df[,4:6] >2] <- NA
 df$count <- rowSums(indx,na.rm=TRUE)
 df
 #   id_A id_B id_C calc_A calc_B calc_C count
 #1   x,z    d  g,f      1      1      5     3
 #2 x,y,z  d,e    f      1      2      8     5
 #3   y,z  d,e    g      6      7      1     1

Update

Suppose, your dataset is not in the same order

假设您的数据集不是按相同的顺序排列的

 set.seed(42)
 df1 <- df[,sample(6)]
 library(gtools)
 df2 <-df1[,mixedorder(names(df1))]
 #    calc_A calc_B calc_C  id_A id_B id_C
 #1      1      1      5   x,z    d  g,f
 #2      1      2      8 x,y,z  d,e    f
 #3      6      7      1   y,z  d,e    g

 id1 <- grep("^id", colnames(df2))
 calc1 <- grep("^calc", colnames(df2)) 

 indx1 <-sapply(df2[, id1], str_count, ",")+1
 indx1[df2[, calc1] >2] <- NA
 df1$count <- rowSums(indx1, na.rm=TRUE)
 df1
 #     calc_C calc_B id_B id_C calc_A  id_A count
 #1      5      1    d  g,f      1   x,z     3
 #2      8      2  d,e    f      1 x,y,z     5
 #3      1      7  d,e    g      6   y,z     1

data

df <- structure(list(id_A = c("x,z", "x,y,z", "y,z"), id_B = c("d", 
 "d,e", "d,e"), id_C = c("g,f", "f", "g"), calc_A = c(1L, 1L, 
 6L), calc_B = c(1L, 2L, 7L), calc_C = c(5L, 8L, 1L)), .Names = c("id_A", 
"id_B", "id_C", "calc_A", "calc_B", "calc_C"), class = "data.frame", row.names = c("1", 
"2", "3"))

#3

I don't know if this is less messy than jbaums solution but here is another option :

我不知道这是否比jbaums解决方案简单，但这里有另一个选择:

mydf<-data.frame(id_A=c("x,y","x,y,z","y,z"),id_B=c("d","d,e","d,e"),id_C=c("g,f","f","g"),
                 calc_A=c(1,1,6),calc_B=c(1,2,7),calc_C=c(5,8,1),stringsAsFactors=F)



mydf$count<-apply(mydf,1,function(rg,namesrg){
                     rg_calc<-rg[grep("calc",namesrg)]
                     rg_ids<-rg[grep("id",namesrg)]
                     idsinf2<-which(as.numeric( rg_calc)<=2)
                     ttids<-unlist(sapply(rg_ids[gsub("calc","id",names(rg_calc[idsinf2]))],function(id){strsplit(id,",")[[1]]}))
                     return(length(ttids))
                    },colnames(mydf))


>  mydf
   id_A id_B id_C calc_A calc_B calc_C count
1   x,y    d  g,f      1      1      5     3
2 x,y,z  d,e    f      1      2      8     5
3   y,z  d,e    g      6      7      1     1