如何在R中的一列中添加具有不同值的新行

Posted 2021-04-25

基本上我有一个所有名称的矢量names，以及一个带有BIN（0/1）字段和NAME字段的数据帧df。对于BIN==0的每一行，我想创建一个重复的行，但是用1代替，并将其添加到df的底部，并使用不同的名称。这是我必须选择一个新名称，给定当前名称：

sample(names[names!=name], 1)

但我不知道如何对此进行矢量化，并使用来自BIN的相同数据将其添加到df。

编辑：示例数据：

df = data.frame(BIN=c(1,0,1), NAME=c("alice","bob","cate"))
names = c("alice","bob","cate","dan")

我接近这样的事情：

rbind(df, df %>% filter(BIN == 0) %>%
    mutate(NAME = sample(names[names!=NAME],1)))

但是我得到一个错误：在binattr（e1，e2）中：长度（e1）不是长度的倍数（e2）。

答案

这是一个简单的方法。我认为这很简单，如果您有疑问，请告诉我：

rename = subset(df, BIN == 0)
rename$NEW_NAME = sample(names, size = nrow(rename), replace = TRUE)
while(any(rename$NAME == rename$NEW_NAME)) {
  matches = rename$NAME == rename$NEW_NAME
  rename$NEW_NAME[matches] = sample(names, size = sum(matches), replace = TRUE)
}
rename$BIN = 1
rename$NAME = rename$NEW_NAME
rename$NEW_NAME = NULL

result = rbind(df, rename)
result
#    BIN  NAME
# 1    1 alice
# 2    0   bob
# 3    1  cate
# 21   1 alice

这是另一种方法，不太清晰但效率更高。这是“正确”的方式，但它需要更多的思考和解释。

df$NAME = factor(df$NAME, levels = names)
rename = subset(df, BIN == 0)
n = length(names)
# we will increment each level number with a random integer from
# 1 to n - 1 (with a mod to make it cyclical)
offset = sample(1:(n - 1), size = nrow(rename), replace = TRUE)
adjusted = (as.integer(rename$NAME) + offset) %% n
# reconcile 1-indexed factor levels with 0-indexed mod operator
adjusted[adjusted == 0] = n
rename$NAME = names[adjusted]
rename$BIN = 1
result = rbind(df, rename)

（或者，为dplyr改写）

df = mutate(df, NAME = factor(NAME, levels = names))
n = length(names)
df %>% filter(BIN == 0) %>%
  mutate(
    offset = sample(1:(n - 1), size = n(), replace = TRUE),
    adjusted = (as.integer(NAME) + offset) %% n,
    adjusted = if_else(adjusted == 0, n, adjusted),
    NAME = names[adjusted],
    BIN = 1
  ) %>%
  select(-offset, -adjusted) %>%
  rbind(df, .)

由于您的问题是矢量化部分，我建议在具有多个BIN 0行的示例案例上测试答案，我使用了这个：

df = data.frame(BIN=c(1,0,1,0,0,0,0,0,0), NAME=rep(c("alice","bob","cate"), 3))

---

而且，因为我很好奇，这里有10个行的基准，有26个名字。结果首先，代码如下：

# Unit: milliseconds
#             expr        min         lq      mean     median         uq        max neval
#       while_loop  34.070438  34.327020  37.53357  35.548047  39.922918  46.206454    10
#        increment   1.397617   1.458592   1.88796   1.526512   2.123894   3.196104    10
#  increment_dplyr  24.002169  24.681960  25.50568  25.374429  25.750548  28.054954    10
#         map_char 346.531498 347.732905 361.82468 359.736403 374.648635 383.575265    10

“聪明”的方式是迄今为止最快的方式。我的猜测是dplyr减速是因为我们不能直接替换adjusted的相关位，而是必须添加if_else的开销。那我们实际上是在adjusted和offset的数据框中添加列而不是处理向量。这足以使它几乎与while循环方法一样慢，这仍然比map_chr快10倍，nn = 10000 df = data.frame( BIN = sample(0:1, size = nn, replace = TRUE, prob = c(0.7, 0.3)), NAME = factor(sample(letters, size = nn, replace = TRUE), levels = letters) ) get.new.name <- function(c){ return(sample(names[names!=c],1)) } microbenchmark::microbenchmark( while_loop = { rename = subset(df, BIN == 0) rename$NEW_NAME = sample(names, size = nrow(rename), replace = TRUE) while (any(rename$NAME == rename$NEW_NAME)) { matches = rename$NAME == rename$NEW_NAME rename$NEW_NAME[matches] = sample(names, size = sum(matches), replace = TRUE) } rename$BIN = 1 rename$NAME = rename$NEW_NAME rename$NEW_NAME = NULL result = rbind(df, rename) }, increment = { rename = subset(df, BIN == 0) n = length(names) # we will increment each level number with a random integer from # 1 to n - 1 (with a mod to make it cyclical) offset = sample(1:(n - 1), size = nrow(rename), replace = TRUE) adjusted = (as.integer(rename$NAME) + offset) %% n # reconcile 1-indexed factor levels with 0-indexed mod operator adjusted[adjusted == 0] = n rename$NAME = names[adjusted] rename$BIN = 1 }, increment_dplyr = { n = length(names) df %>% filter(BIN == 0) %>% mutate( offset = sample(1:(n - 1), size = n(), replace = TRUE), adjusted = (as.integer(NAME) + offset) %% n, adjusted = if_else(adjusted == 0, n, adjusted), NAME = names[adjusted], BIN = 1 ) %>% select(-offset,-adjusted) }, map_char = { new.df <- df %>% filter(BIN == 0) %>% mutate(NAME = map_chr(NAME, get.new.name)) %>% mutate(BIN = 1) }, times = 10 ) 必须一次排成一排。

library(tidyverse)

df <- data.frame(BIN=c(1,0,1), NAME=c("alice","bob","cate"), stringsAsFactors = FALSE)
names <- c("alice","bob","cate","dan")

df %>% 
  mutate(NAME_new = ifelse(BIN == 0, sample(names, n(), replace = TRUE), NA)) %>% 
  gather(name_type, NAME, NAME:NAME_new, na.rm = TRUE) %>% 
  mutate(BIN = ifelse(name_type == "NAME_new", 1, BIN)) %>% 
  select(-name_type)

另一答案

有点奇怪，但我认为这应该是你想要的：

  BIN  NAME
1   1 alice
2   0   bob
3   1  cate
4   1 alice

输出：

rowwise()

另一答案

好吧，我不打算回答我自己的问题，但我确实找到了一个更简单的解决方案。我认为它比使用library(tidyverse) get.new.name <- function(c){ return(sample(names[names!=c],1)) } new.df <- rbind(df, df %>% filter(BIN == 0) %>% mutate(NAME = map_chr(NAME, get.new.name)) %>% mutate(BIN = 1) 更好，但我不知道它是否必然是最有效的方式。

map_char

map最终变得非常重要，而不仅仅是qazxswpoi，因为后者将返回一个奇怪的列表列表。