如何在dplyr管线中使用sample和seq？

问题描述

我有一个包含两列的数据框，低和高。我想使用dplyr创建一个新变量，该变量是一个介于低值和高值（包含和相等的概率）之间的随机选择的值。我已经尝试过

I have a dataframe with two columns, low and high. I would like to create a new variable that is a randomly selected value between low and high (inclusive and equal probability) using dplyr. I have tried

library(tidyverse)

data_frame(low = 1:10, high = 11) %>% 
    mutate(rand_btwn = base::sample(seq(low, high, by = 1), size = 1))

这会给我一个错误，因为 seq 需要标量参数。

which gives me an error since seq expects scalar arguments.

然后我再次尝试使用矢量化版本的 seq

I then tried again using a vectorized version of seq

seq2 <- Vectorize(seq.default, vectorize.args = c("from", "to"))

data_frame(low = 1:10, high = 11) %>% 
    mutate(rand_btwn = base::sample(seq2(low, high, by = 1), size = 1))

但这也不能给我想要的结果。

but this does not give me the desired result either.

推荐答案

要避免 rowwise（）模式，我通常更喜欢 mutate（）中的 map（） ，例如：

To avoid the rowwise() pattern, I usually prefer to map() in mutate(), like:

set.seed(123)
data_frame(low = 1:10, high = 11) %>% 
  mutate(rand_btwn = map_int(map2(low, high, seq), sample, size = 1))
# # A tibble: 10 x 3
#      low  high rand_btwn
#    <int> <dbl>     <int>
#  1     1    11         4
#  2     2    11         9
#  3     3    11         6
#  4     4    11        11
#  5     5    11        11
#  6     6    11         6
#  7     7    11         9
#  8     8    11        11
#  9     9    11        10
# 10    10    11        10

或：

set.seed(123)
data_frame(low = 1:10, high = 11) %>% 
  mutate(rand_btwn = map2_int(low, high, ~ sample(seq(.x, .y), 1)))

您的 Vectorize（）方法也适用：

sample_v <- Vectorize(function(x, y) sample(seq(x, y), 1))

set.seed(123)
data_frame(low = 1:10, high = 11) %>% 
  mutate(rand_btwn = sample_v(low, high))

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