R lead或lag函数可获得多个值，而不仅仅是第n个值
我有一个tibble，每行有一个单词列表。我想从一个搜索关键字的函数中创建一个新变量，如果它找到关键字，则创建一个由关键字加减3个单词组成的字符串 下面的代码很接近，但是，它没有抓住我的关键字前后的所有三个单词，而是抓住前面/后面的单个单词3R lead或lag函数可获得多个值，而不仅仅是第n个值,r,dplyr,lag,lead,R,Dplyr,Lag,Lead,我有一个tibble，每行有一个单词列表。我想从一个搜索关键字的函数中创建一个新变量，如果它找到关键字，则创建一个由关键字加减3个单词组成的字符串 下面的代码很接近，但是，它没有抓住我的关键字前后的所有三个单词，而是抓住前面/后面的单个单词3 df <- tibble(words = c("it", "was", "the", "best", "of", "times", "it", "was", "the", "worst", "of",
df <- tibble(words = c("it", "was", "the", "best", "of", "times",
"it", "was", "the", "worst", "of", "times"))
df <- df %>% mutate(chunks = ifelse(words=="times",
paste(lag(words, 3),
words,
lead(words, 3), sep = " "),
NA))
df一个选项是sapply
：
library(dplyr)
df %>%
mutate(
chunks = ifelse(
words == "times",
sapply(
1:nrow(.),
function(x) paste(words[pmax(1, x - 3):pmin(x + 3, nrow(.))], collapse = " ")
),
NA
)
)
输出：
# A tibble: 12 x 2
words chunks
<chr> <chr>
1 it NA
2 was NA
3 the NA
4 best NA
5 of NA
6 times the best of times it was the
7 it NA
8 was NA
9 the NA
10 worst NA
11 of NA
12 times the worst of times
#一个tible:12 x 2
词块
1 it NA
2是NA
3北美
4最佳北美酒店
NA的5
6次最好的时候是
7它是NA
8是NA
9北美
10最差的NA
11/NA
是最糟糕的12倍
虽然不是一个明确的lead
或lag
功能，但它通常也可以用于此目的。类似于@arg0naut，但没有dplyr：
r = 1:nrow(df)
w = which(df$words == "times")
wm = lapply(w, function(wi) intersect(r, seq(wi-3L, wi+3L)))
df$chunks <- NA_character_
df$chunks[w] <- tapply(df$words[unlist(wm)], rep(w, lengths(wm)), FUN = paste, collapse=" ")
# A tibble: 12 x 2
words chunks
<chr> <chr>
1 it <NA>
2 was <NA>
3 the <NA>
4 best <NA>
5 of <NA>
6 times the best of times it was the
7 it <NA>
8 was <NA>
9 the <NA>
10 worst <NA>
11 of <NA>
12 times the worst of times
data.table:：shift
为n
（lag）参数接受一个向量并输出一个列表，因此您可以使用该向量和do.call（将列表元素粘贴在一起）。但是，除非您使用的是data.table version>=1.12，否则我认为它不会让您混合正负n
值（如下所示）
使用数据表：
library(data.table)
setDT(df)
df[, chunks := trimws(ifelse(words != "times", NA, do.call(paste, shift(words, 3:-3, ''))))]
# words chunks
# 1: it <NA>
# 2: was <NA>
# 3: the <NA>
# 4: best <NA>
# 5: of <NA>
# 6: times the best of times it was the
# 7: it <NA>
# 8: was <NA>
# 9: the <NA>
# 10: worst <NA>
# 11: of <NA>
# 12: times the worst of times
下面是另一个使用lag
和lead
laglead_f <- function(what, range)
setNames(paste(what, "(., ", range, ", default = '')"), paste(what, range))
df %>%
mutate_at(vars(words), funs_(c(laglead_f("lag", 3:0), laglead_f("lead", 1:3)))) %>%
unite(chunks, -words, sep = " ") %>%
mutate(chunks = ifelse(words == "times", trimws(chunks), NA))
## A tibble: 12 x 2
# words chunks
# <chr> <chr>
# 1 it NA
# 2 was NA
# 3 the NA
# 4 best NA
# 5 of NA
# 6 times the best of times it was the
# 7 it NA
# 8 was NA
# 9 the NA
#10 worst NA
#11 of NA
#12 times the worst of times
laglead\u f%
变异（变量（单词），funs（滞后，3:0），滞后（1:3）））%>%
联合（块，-words，sep=“”）%>%
变异（块=ifelse（单词=times），trimws（块），NA））
##一个tibble:12x2
#词块
#
#1 it NA
#2是NA
#3北美
#4最佳北美酒店
#NA的5
#6次最好的时候是
#7它是NA
#8是NA
#9北美
#10最差的NA
#11/NA
#是最糟糕的12倍
其思想是将三个lag
ged和lead
ing向量中的值存储在新的列中，使用mutate\u at
和一个命名函数unite
这些列，然后根据您的条件过滤条目，其中words==“times”
library(dplyr)
df %>%
mutate(chunks = do.call(paste, data.table::shift(words, 3:-3, fill = '')),
chunks = trimws(ifelse(words != "times", NA, chunks)))
# # A tibble: 12 x 2
# words chunks
# <chr> <chr>
# 1 it NA
# 2 was NA
# 3 the NA
# 4 best NA
# 5 of NA
# 6 times the best of times it was the
# 7 it NA
# 8 was NA
# 9 the NA
# 10 worst NA
# 11 of NA
# 12 times the worst of times
laglead_f <- function(what, range)
setNames(paste(what, "(., ", range, ", default = '')"), paste(what, range))
df %>%
mutate_at(vars(words), funs_(c(laglead_f("lag", 3:0), laglead_f("lead", 1:3)))) %>%
unite(chunks, -words, sep = " ") %>%
mutate(chunks = ifelse(words == "times", trimws(chunks), NA))
## A tibble: 12 x 2
# words chunks
# <chr> <chr>
# 1 it NA
# 2 was NA
# 3 the NA
# 4 best NA
# 5 of NA
# 6 times the best of times it was the
# 7 it NA
# 8 was NA
# 9 the NA
#10 worst NA
#11 of NA
#12 times the worst of times