问题
I have a column f in my dataframe that I would like to spread into multiple columns based on the values in that column. For example:
df <- structure(list(f = c(NA, "18,17,10", "12,8", "17,11,6", "18",
"12", "12", NA, "17,11", "12")), .Names = "f", row.names = c(NA,
10L), class = "data.frame")
df
# f
# 1 <NA>
# 2 18,17,10
# 3 12,8
# 4 17,11,6
# 5 18
# 6 12
# 7 12
# 8 <NA>
# 9 17,11
# 10 12
How would I split column f into multiple columns indicating the numbers in the row. I'm interested in something like this:
6 8 10 11 12 17 18
1 0 0 0 0 0 0 0
2 0 0 1 0 0 1 1
3 0 1 0 0 1 0 0
4 1 0 0 1 0 1 0
5 0 0 0 0 0 0 1
6 0 0 0 0 1 0 0
7 0 0 0 0 1 0 0
8 0 0 0 0 0 0 0
9 0 0 0 1 0 1 0
10 0 0 0 0 1 0 0
I'm thinking I could useunique on the f column to create the seperate columns based on the different numbers and then do a grepl to determine if the specific number is in column f but I was wondering if there was a better way. Something similar to spread or separate in the tidyr package.
回答1:
A solution using tidyr::separate_rows will be as:
library(tidyverse)
df %>% mutate(ind = row_number()) %>%
separate_rows(f, sep=",") %>%
mutate(f = ifelse(is.na(f),0, f)) %>%
count(ind, f) %>%
spread(f, n, fill = 0) %>%
select(-2) %>% as.data.frame()
# ind 10 11 12 17 18 6 8
# 1 1 0 0 0 0 0 0 0
# 2 2 1 0 0 1 1 0 0
# 3 3 0 0 1 0 0 0 1
# 4 4 0 1 0 1 0 1 0
# 5 5 0 0 0 0 1 0 0
# 6 6 0 0 1 0 0 0 0
# 7 7 0 0 1 0 0 0 0
# 8 8 0 0 0 0 0 0 0
# 9 9 0 1 0 1 0 0 0
# 10 10 0 0 1 0 0 0 0
回答2:
This could be achieved by splitting on the ,, the stack it to a two column data.frame and get the frequency with table
df1 <- na.omit(stack(setNames(lapply(strsplit(df$f, ","),
as.numeric), seq_len(nrow(df))))[, 2:1])
table(df1)
# values
#ind 6 8 10 11 12 17 18
# 1 0 0 0 0 0 0 0
# 2 0 0 1 0 0 1 1
# 3 0 1 0 0 1 0 0
# 4 1 0 0 1 0 1 0
# 5 0 0 0 0 0 0 1
# 6 0 0 0 0 1 0 0
# 7 0 0 0 0 1 0 0
# 8 0 0 0 0 0 0 0
# 9 0 0 0 1 0 1 0
# 10 0 0 0 0 1 0 0
来源：https://stackoverflow.com/questions/50650152/split-column-of-comma-separated-numbers-into-multiple-columns-based-on-value