Question

我有以下带有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L中给出的值将第2列和第3列分组：

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，我希望获得两个新列，第一个称为intervals_b，它指示列b（第2列）的值是否在-Inf和-0.3之间，- 0.3和0.7或0.7和Inf，对于c列也类似：-Inf到-1，-1到0.5、0.5到1和1到Inf。

我遇到错误：

mutate_impl（.data，点）中的错误：评估错误：递归索引在第2级失败

我想在一般情况下这样做，这就是为什么我使用隐式名称。

有什么想法吗？

Answer 1

您可以通过并行传递mapply的{{1}}和ColsToChoose的{{1}}来执行此基数R df，以获得范围。

使用相同方法的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]方法

tidyverse

这将提供与上面相同的输出。

基于列表中的限制的切割间隔

1 个答案: