我有以下包含 4 个数字列的数据框:
df <- structure(list(a = c(0.494129340746821, 1.0182303327812, 0.412227511922328,
0.204436644926016, 0.707038309818134, -0.0547300783473556, 1.02124944293185,
0.381284586356091, 0.375197843213519, -1.18172401075089), b =
c(-1.34374367808722,
-0.724644569211516, -0.618107980582741, -1.79274868750102,
-3.03559838445132,
-0.205726144151615, -0.441511286334811, 0.126660637747845,
0.353737902975931,
-0.26601393471207), c = c(1.36922677098999, -1.81698348029464,
-0.846111260721092, 0.121256015837603, -1.16499681749603, 1.14145675696301,
-0.782988942359773, 3.25142254765012, -0.132099541183856, -0.242831877642412
), d = c(-0.30002630673509, -0.507496812070994, -2.59870853299723,
-1.30109828239028, 1.05029458887117, -0.606381379180569, -0.928822706709913,
-0.68324741261771, -1.17980245487707, 2.20174180936794)), row.names = c(NA,
-10L), class = c("tbl_df", "tbl", "data.frame"))
我想创建两个新的因子列,其中我根据列表中给出的值对第 2 列和第 3 列进行分组 L
:
ColsToChoose = c(2,3)
L = list()
L[[1]] = c(-0.3, 0.7)
L[[2]] = c(-1, 0.5, 1)
df %>% mutate_at(ColsToChoose, funs(intervals = cut(., c(-Inf, L[[.]], Inf))))
也就是说,我期望得到两个新列,第一列称为 intervals_b
指示第 b
列(第 2 列)的值是否在 -Inf
和 -0.3、-0.3 和 0.7 或 0.7 和 Inf
之间,同样对于列 c
:-Inf
到 -1、-1 到 0.5、0.5 到 1 和 1 到 Inf
之间。
我收到一个错误:
错误 mutate_impl(.data, dots) : 评估错误:递归索引在级别 2 失败
我想在一般情况下这样做,这就是我使用隐式名称的原因。
有什么想法吗?
你可以mapply
传递df
的ColsToChoose
并L
并行来获得这个基础 R,以获得范围。
df[paste0("interval", names(df)[ColsToChoose])] <-
mapply(function(x, y) cut(x, c(-Inf, y, Inf)), df[ColsToChoose], L)
df
# a b c d intervalb intervalc
# <dbl> <dbl> <dbl> <dbl> <chr> <chr>
# 1 0.494 -1.34 1.37 -0.300 (-Inf,-0.3] (1, Inf]
# 2 1.02 -0.725 -1.82 -0.507 (-Inf,-0.3] (-Inf,-1]
# 3 0.412 -0.618 -0.846 -2.60 (-Inf,-0.3] (-1,0.5]
# 4 0.204 -1.79 0.121 -1.30 (-Inf,-0.3] (-1,0.5]
# 5 0.707 -3.04 -1.16 1.05 (-Inf,-0.3] (-Inf,-1]
# 6 -0.0547 -0.206 1.14 -0.606 (-0.3,0.7] (1, Inf]
# 7 1.02 -0.442 -0.783 -0.929 (-Inf,-0.3] (-1,0.5]
# 8 0.381 0.127 3.25 -0.683 (-0.3,0.7] (1, Inf]
# 9 0.375 0.354 -0.132 -1.18 (-0.3,0.7] (-1,0.5]
#10 -1.18 -0.266 -0.243 2.20 (-0.3,0.7] (-1,0.5]
使用相同方法的tidyverse
方法
library(tidyverse)
bind_cols(df,
map2(df[ColsToChoose], L, ~ cut(.x, c(-Inf, .y, Inf))) %>%
data.frame() %>%
rename_all(paste0, "_interval"))
这给出了与上述相同的输出。