Question

我已对limesurvey进行了调查，并将结果导出为csv.file，并将其导入R。

其中一个问题是多项选择题，参与者可以在其中选择自己所学的科目。 limesurvey的输出看起来像这样（但是有更多的主题和更多的参与者）：

Participant | Maths | Physics | English | Biology 
1           |   Y   |         |    Y    |         
2           |       |    Y    |    Y    |         
3           |   Y   |    Y    |         |   Y

我想要一个看起来像这样的结果

Participant | Subject 1 | Subject 2| Subject 3  |
1           |   Maths   | English  |            |         
2           |   Physics | English  |            |         
3           |   Maths   | Physics  | Biology    |

对于任何指针，我将不胜感激。

Answer 1

这是我尝试按照要求生成预期数据帧的方法：

library(tidyverse)
library(gtools)
rand_list = c('Y', NA)
df = data.frame(participant = seq(1,10, by = 1), # r starts counting from 0
                Maths = sample(rand_list, 10, replace = TRUE),
                Physics = sample(rand_list, 10, replace = TRUE),
                English = sample(rand_list, 10, replace = TRUE),
                Biology = sample(rand_list, 10, replace = TRUE))

df_to_new_format = function(data){
  vector_subject = colnames(data)
  vector_new_col = c()
  for (i in 1:length(vector_subject)){
    if (i == 1){
      new_col = 'participant'
      vector_new_col <- c(vector_new_col, new_col)
      rm(new_col)
    } else{
      new_col = paste('Subject', as.character(i - 1))
      vector_new_col <- c(vector_new_col, new_col)
      rm(new_col)
    }
  }

  for (j in 1:length(vector_subject)){
    if (j == 1){
      next
    } else{
      data[[j]] <- recode(data[[j]], 'Y' = vector_subject[j])
    }
  }

  colnames(data) <- vector_new_col
  return(data)
}

df = df_to_new_format(data = df)
df_new_format = c()

for (m in 1:nrow(df)){
  temp = mixedsort(as.matrix(df[m,]))
  print(temp)
  df_new_format = rbind(df_new_format, temp)
}

df_new_format = as.data.frame(df_new_format, row.names = FALSE)
colnames(df_new_format) = colnames(df)

Answer 2

我对这种数据争执有点不习惯，但是这里有一些建议。

首先让我们假设您的数据具有以下格式：

dtf <- structure(list(Participant = c("1", "2", "3", "4"),
Physics = c("Y", "Y", "N", "N"), Chemistry = c("Y", "N", "N",
"N"), Math = c("N", "Y", "Y", "Y"), Biology = c("N", "Y", "N",
"Y")), class = "data.frame", row.names = c(NA, -4L))

然后我们可以重新安排这样的事情

wh <- which(dtf == "Y", arr.ind=TRUE)
tapply(wh[,2], wh[,1], function(x) colnames(dtf)[x])
# $`1`
# [1] "Physics"   "Chemistry"

# $`2`
# [1] "Physics" "Math"    "Biology"

# $`3`
# [1] "Math"

# $`4`
# [1] "Math"    "Biology"

或

dtf2 <- dtf[1]
dtf2$Subject <- apply(dtf, 1, function(r) {c(names(dtf)[r == "Y"])})
dtf2
#   Participant                Subject
# 1           1     Physics, Chemistry
# 2           2 Physics, Math, Biology
# 3           3                   Math
# 4           4          Math, Biology

或者使用melt()中的dcast()和reshape2

library(reshape2)

dtf.m <- melt(dtf, 1)
dcast(dtf.m[dtf.m$value == "Y", 1:2], Participant ~ variable)
#   Participant Physics Chemistry Math Biology
# 1           1 Physics Chemistry <NA>    <NA>
# 2           2 Physics      <NA> Math Biology
# 3           3    <NA>      <NA> Math    <NA>
# 4           4    <NA>      <NA> Math Biology

重新编码调查选择题的输出

2 个答案: