Question

我的数据有受访者（本例中为10个），可在n个选项中选择（本例中为3个）。

# original data
df <- data_frame(RID = seq(1:10), choice = sample(1:3,10,replace = TRUE))

我正在尝试将其编码为二进制值，但使用dplyr进行长（整）格式编码。我的狡猾感告诉我，可能有一种比使用spread和gather更好的方法。

# desired output
df %>%
  mutate(value = 1) %>%
  spread(choice,value, fill=0) %>%
  gather("choice","selection",2:4) %>%
  arrange(RID,choice)

有关更好方法的任何想法吗？

Answer 1

使用tidyr::complete从列中创建唯一值的所有组合（此处您需要RID和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

Answer 2

另一种选择可能是使用expand.grid：

#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

dplyr中长格式的二进制编码

2 个答案: