Frequency counts in R [duplicate]

2019-01-01 13:45发布

This question already has an answer here:

This may seem like a very basic R question, but I'd appreciate an answer. I have a data frame in the form of:

col1    col2
a   g
a   h
a   g
b   i
b   g
b   h
c   i

I want to transform it into counts, so the outcome would be like this. I've tried using table () function, but seem to only be able to get the count for one column.

    a   b   c
g   2   1   0
h   1   1   0
i   0   1   1

How do I do it in R?

2条回答
残风、尘缘若梦
2楼-- · 2019-01-01 14:13

Using f from @Ananda you can use dcast

library(reshape2)

> dcast(f, V1~V2)
Using V2 as value column: use value.var to override.
Aggregation function missing: defaulting to length
  V1  g  h  i
1 a   2  1  0
2 b   1  1  1
3 c   0  0  1

However, I'm writing this only in case you may need something more than just table (which for this case it's the simplest correct answer) in the future, like:

set.seed(1)
f$var <- rnorm(7)

> f
  V1 V2        var
1 a   g -0.6264538
2 a   h  0.1836433
3 a   g -0.8356286
4 b   i  1.5952808
5 b   g  0.3295078
6 b   h -0.8204684
7 c   i  0.4874291

> dcast(f, V1~V2, value.var="var", fun.aggregate=sum)
  V1          g          h         i
1 a  -1.4620824  0.1836433 0.0000000
2 b   0.3295078 -0.8204684 1.5952808
3 c   0.0000000  0.0000000 0.4874291
查看更多
爱死公子算了
3楼-- · 2019-01-01 14:30

I'm not really sure what you used, but table works fine for me!

Here's a minimal reproducible example:

df <- structure(list(V1 = c("a", "a", "a", "b", "b", "b", "c"), 
                     V2 = c("g", "h", "g", "i", "g", "h", "i")), 
                .Names = c("V1", "V2"), class = "data.frame", 
                row.names = c(NA, -7L))
table(df)
#    V2
# V1  g h i
#   a 2 1 0
#   b 1 1 1
#   c 0 0 1

Notes:

  • Try table(df[c(2, 1)]) (or table(df$V2, df$V1)) to swap the rows and columns.
  • Use as.data.frame.matrix(table(df)) to get a data.frame as your output. (as.data.frame will create a long data.frame, not one in the same output format you desire).
查看更多
登录 后发表回答