dplyr中的长格式二进制编码

我的数据包含在n个选项（本例中为3个）之间进行选择的受访者（本例中为10个）

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有什么更好的方法吗？

使用

tidyr:：complete

从列中创建所有唯一值的组合（这里您需要

RID

和

choice

）：

df%>%
mutate（selection=1）%>%#创建一个1的选择列
完成（RID，choice，fill=list（selection=0））#用0填充缺少的组合
#一个tibble:30x3
#RID选择
#          
# 1     1      1        1.
# 2     1      2        0.
# 3     1      3        0.
# 4     2      1        0.
# 5     2      2        0.
# 6     2      3        1.
# 7     3      1        0.
# 8     3      2        0.
# 9     3      3        1.
#10     4      1        1.
# ... 还有20行

另一个选项可以使用

展开.grid

作为：

#Create all possible combination using RID and unique choice
result <- expand.grid(RID = df$RID, choice = unique(df$choice))

#New column as 'select' will be 1 for those combination which were present in original df
result$selection = ifelse(result$RID == df$RID & result$choice == df$choice, 1, 0)

result

#1    1      2         1
#2    2      2         0
#3    3      2         0
#4    4      2         0
#5    5      2         0
#6    6      2         0
#7    7      2         0
#8    8      2         0
#9    9      2         1
#........
#........
#30 rows

#使用RID和unique choice创建所有可能的组合
结果
# desired output
df %>%
  mutate(value = 1) %>%
  spread(choice,value, fill=0) %>%
  gather("choice","selection",2:4) %>%
  arrange(RID,choice)

df %>% 
    mutate(selection = 1) %>%                         # create a selection column of 1
    complete(RID, choice, fill = list(selection = 0)) # fill selection with 0 for missing combinations

# A tibble: 30 x 3
#     RID choice selection
#   <int>  <int>     <dbl>
# 1     1      1        1.
# 2     1      2        0.
# 3     1      3        0.
# 4     2      1        0.
# 5     2      2        0.
# 6     2      3        1.
# 7     3      1        0.
# 8     3      2        0.
# 9     3      3        1.
#10     4      1        1.
# ... with 20 more rows

#Create all possible combination using RID and unique choice
result <- expand.grid(RID = df$RID, choice = unique(df$choice))

#New column as 'select' will be 1 for those combination which were present in original df
result$selection = ifelse(result$RID == df$RID & result$choice == df$choice, 1, 0)

result

#1    1      2         1
#2    2      2         0
#3    3      2         0
#4    4      2         0
#5    5      2         0
#6    6      2         0
#7    7      2         0
#8    8      2         0
#9    9      2         1
#........
#........
#30 rows