Making a character string with column names with zero values

隐身守侯 提交于 2019-12-31 04:28:05

问题


The 4th column is my desired column. Video,Webinar,Meeting,Conference are the 4 type of activities that the different customers(names) can engage in. You can see,in a given row, all the column names with zero value are in the final column(NextStep) and the value there(character string separated by commas) excludes the column name with non-zero value. The character strings(column names) in the final column usually appear in the column order with two exceptions. Webinar always appears first if it has a zero value and video always appears last if it has a zero value.

    library(data.table)
     dt <- fread('
 Name     Video   Webinar Meeting Conference   NextStep
  John       1         0        0       0         Webinar,Meeting,Conference
  John       1         1        0       0         Meeting,Conference
  John       1         1        1       0         Conference      
  Tom        0         0        1       0         Webinar,Conference,Video
  Tom        0         0        1       1         Webinar,Video   
  Kyle       0         0        0       1         Webinar,Meeting,Video

                                    ')

My question is how to create the next step column. Thanks a lot for your help!


回答1:


A possible solution:

DT[, nextstep := paste0(names(.SD)[.SD==0], collapse = ','), 1:nrow(DT), .SDcols = 2:5][]

which gives:

   Name Video Webinar Meeting Conference                   nextstep
1: John     1       0       0          0 Webinar,Meeting,Conference
2: John     1       1       0          0         Meeting,Conference
3: John     1       1       1          0                 Conference
4:  Tom     0       0       1          0   Video,Webinar,Conference
5:  Tom     0       0       1          1              Video,Webinar
6: Kyle     0       0       0          1      Video,Webinar,Meeting

When you want to order the names as you specified in the comments, you can do:

lvls <- c('Webinar', 'Meeting', 'Conference', 'Video')
DT[, nextstep := paste0(lvls[lvls %in% names(.SD)[.SD==0]], collapse = ','), 
   1:nrow(DT), .SDcols = 2:5][]

which gives:

   Name Video Webinar Meeting Conference                   nextstep
1: John     1       0       0          0 Webinar,Meeting,Conference
2: John     1       1       0          0         Meeting,Conference
3: John     1       1       1          0                 Conference
4:  Tom     0       0       1          0   Webinar,Conference,Video
5:  Tom     0       0       1          1              Webinar,Video
6: Kyle     0       0       0          1      Webinar,Meeting,Video

Instead of using paste0 (with collapse = ',') you can also use toString.


Used data:

DT <- fread('Name     Video   Webinar  Meeting  Conference
             John       1         0        0        0
             John       1         1        0        0
             John       1         1        1        0
             Tom        0         0        1        0
             Tom        0         0        1        1
             Kyle       0         0        0        1')



回答2:


In case you are looking for a way to do this without simply re-ordering the columns in the order you want (in fact I see no reason why not to do so, but anyway..) you could try the following approach. It melts and updates by reference in a join:

lvls <- c("Webinar", "Meeting", "Conference", "Video")  # make sure order is correct
dt[, row := .I]   # add a row-identifier
dtm <- melt(dt, id.vars = c("Name", "row"), measure.vars = lvls) # melt to long format
# summarise dtm by using factor, sorting it and converting to strin; then join to dt
dt[dtm[value == 0, list(NextStep2 = toString(sort(factor(variable, levels = lvls)))), 
    by = row], NextStep2 := NextStep2, on = "row"][, row := NULL]

#    Name Video Webinar Meeting Conference                   NextStep                    NextStep2
# 1: John     1       0       0          0 Webinar,Meeting,Conference Webinar, Meeting, Conference
# 2: John     1       1       0          0         Meeting,Conference          Meeting, Conference
# 3: John     1       1       1          0                 Conference                   Conference
# 4:  Tom     0       0       1          0   Webinar,Conference,Video   Webinar, Conference, Video
# 5:  Tom     0       0       1          1              Webinar,Video               Webinar, Video
# 6: Kyle     0       0       0          1      Webinar,Meeting,Video      Webinar, Meeting, Video

If you want to paste all column names as in the data for those cases where there's no activity, you can add the following line to your code:

dt[rowSums(dt[, mget(lvls)]) == 0, NextStep2 := toString(names(dt)[2:5])]



回答3:


Here you go:

setcolorder(dt, c("Name", "Webinar", "Meeting", "Conference", "Video", "NextStep"))
dt[, NextStepNew:=apply(dt, 1, function(x) paste0(names(x)[x==0], collapse=","))][]
   Name Webinar Meeting Conference Video                   NextStep                NextStepNew
1: John       0       0          0     1 Webinar,Meeting,Conference Webinar,Meeting,Conference
2: John       1       0          0     1         Meeting,Conference         Meeting,Conference
3: John       1       1          0     1                 Conference                 Conference
4:  Tom       0       1          0     0   Webinar,Conference,Video   Webinar,Conference,Video
5:  Tom       0       1          1     0              Webinar,Video              Webinar,Video
6: Kyle       0       0          1     0      Webinar,Meeting,Video      Webinar,Meeting,Video


来源:https://stackoverflow.com/questions/37734904/making-a-character-string-with-column-names-with-zero-values

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