grepl() 和 lapply 填充缺失值
•浏览 1
grepl() and lapply to fill missing values
我以以下数据为例：
fruit.region <- data.frame(full =c("US red apple","bombay Asia mango","gold kiwi New Zealand"), name = c("apple","mango","kiwi"), country = c("US","Asia","New Zealand"), type = c("red","bombay","gold"))new.entry <- c("bombay US mango")
split.new.entry <- strsplit(new.entry,"")
lapply(split.new.entry, function(x){
check = grepl(x, fruit.region, ignore.case=TRUE)
print(check)
})library(stringr)
addnewrow <- function(newfruit){
z<-lapply(fruit.region[,2:4], function(x) x[str_detect(new.entry, x)])
z$full <- newfruit
z
}
addnewrow(new.entry)
$name
[1]"mango"
$country
[1]"US"
$type
[1]"bombay"
$full
[1]"bombay US mango"rbind(fruit.region, addnewrow(new.entry))z <- do.call(rbind, lapply(c(new.entry, new.entry), addnewrow))
rbind(fruit.region, z)fruit.region[] <- lapply(fruit.region, as.character)我希望 R 能够查看 "full" (name) 列中没有 "name"、"country" 和 "type\\ 值的其他项目"并查看它们是否与其他项目匹配。例如，如果 full 的第 4 行带有"bombay US mango"，它将能够识别出国家应该读作 US，bombay 应该在 type 下，mango 应该在 name 下。
这是我目前所拥有的，它只是(逻辑上)识别项目匹配的位置：
fruit.region <- data.frame(full =c("US red apple","bombay Asia mango","gold kiwi New Zealand"), name = c("apple","mango","kiwi"), country = c("US","Asia","New Zealand"), type = c("red","bombay","gold"))new.entry <- c("bombay US mango")
split.new.entry <- strsplit(new.entry,"")
lapply(split.new.entry, function(x){
check = grepl(x, fruit.region, ignore.case=TRUE)
print(check)
})library(stringr)
addnewrow <- function(newfruit){
z<-lapply(fruit.region[,2:4], function(x) x[str_detect(new.entry, x)])
z$full <- newfruit
z
}
addnewrow(new.entry)
$name
[1]"mango"
$country
[1]"US"
$type
[1]"bombay"
$full
[1]"bombay US mango"rbind(fruit.region, addnewrow(new.entry))z <- do.call(rbind, lapply(c(new.entry, new.entry), addnewrow))
rbind(fruit.region, z)fruit.region[] <- lapply(fruit.region, as.character)我有点停滞不前..我已经阅读了许多正则表达式帖子和 grepl 上的 r 帮助指南，但无法找到一个很好的解决方案。我所拥有的不能完全识别逻辑"匹配"向量，因此我无法对不同元素进行子集化并使用 if 语句连接。理想情况下，我希望能够以 data.table 形式替换这些元素，因为我的 fruit.region 实际上会在数据表中。有人对最佳方法有任何建议吗？
使用 stringr 库中的 str_detect 函数。这给出了一个列表，准备 rbind:
fruit.region <- data.frame(full =c("US red apple","bombay Asia mango","gold kiwi New Zealand"), name = c("apple","mango","kiwi"), country = c("US","Asia","New Zealand"), type = c("red","bombay","gold"))new.entry <- c("bombay US mango")
split.new.entry <- strsplit(new.entry,"")
lapply(split.new.entry, function(x){
check = grepl(x, fruit.region, ignore.case=TRUE)
print(check)
})library(stringr)
addnewrow <- function(newfruit){
z<-lapply(fruit.region[,2:4], function(x) x[str_detect(new.entry, x)])
z$full <- newfruit
z
}
addnewrow(new.entry)
$name
[1]"mango"
$country
[1]"US"
$type
[1]"bombay"
$full
[1]"bombay US mango"rbind(fruit.region, addnewrow(new.entry))z <- do.call(rbind, lapply(c(new.entry, new.entry), addnewrow))
rbind(fruit.region, z)fruit.region[] <- lapply(fruit.region, as.character)下一步将取决于您想要的结果 - 如果您只想添加一个，请尝试：
fruit.region <- data.frame(full =c("US red apple","bombay Asia mango","gold kiwi New Zealand"), name = c("apple","mango","kiwi"), country = c("US","Asia","New Zealand"), type = c("red","bombay","gold"))new.entry <- c("bombay US mango")
split.new.entry <- strsplit(new.entry,"")
lapply(split.new.entry, function(x){
check = grepl(x, fruit.region, ignore.case=TRUE)
print(check)
})library(stringr)
addnewrow <- function(newfruit){
z<-lapply(fruit.region[,2:4], function(x) x[str_detect(new.entry, x)])
z$full <- newfruit
z
}
addnewrow(new.entry)
$name
[1]"mango"
$country
[1]"US"
$type
[1]"bombay"
$full
[1]"bombay US mango"rbind(fruit.region, addnewrow(new.entry))z <- do.call(rbind, lapply(c(new.entry, new.entry), addnewrow))
rbind(fruit.region, z)fruit.region[] <- lapply(fruit.region, as.character)如果你有很多：
fruit.region <- data.frame(full =c("US red apple","bombay Asia mango","gold kiwi New Zealand"), name = c("apple","mango","kiwi"), country = c("US","Asia","New Zealand"), type = c("red","bombay","gold"))new.entry <- c("bombay US mango")
split.new.entry <- strsplit(new.entry,"")
lapply(split.new.entry, function(x){
check = grepl(x, fruit.region, ignore.case=TRUE)
print(check)
})library(stringr)
addnewrow <- function(newfruit){
z<-lapply(fruit.region[,2:4], function(x) x[str_detect(new.entry, x)])
z$full <- newfruit
z
}
addnewrow(new.entry)
$name
[1]"mango"
$country
[1]"US"
$type
[1]"bombay"
$full
[1]"bombay US mango"rbind(fruit.region, addnewrow(new.entry))z <- do.call(rbind, lapply(c(new.entry, new.entry), addnewrow))
rbind(fruit.region, z)fruit.region[] <- lapply(fruit.region, as.character)请确保您的列是字符优先：
fruit.region <- data.frame(full =c("US red apple","bombay Asia mango","gold kiwi New Zealand"), name = c("apple","mango","kiwi"), country = c("US","Asia","New Zealand"), type = c("red","bombay","gold"))new.entry <- c("bombay US mango")
split.new.entry <- strsplit(new.entry,"")
lapply(split.new.entry, function(x){
check = grepl(x, fruit.region, ignore.case=TRUE)
print(check)
})library(stringr)
addnewrow <- function(newfruit){
z<-lapply(fruit.region[,2:4], function(x) x[str_detect(new.entry, x)])
z$full <- newfruit
z
}
addnewrow(new.entry)
$name
[1]"mango"
$country
[1]"US"
$type
[1]"bombay"
$full
[1]"bombay US mango"rbind(fruit.region, addnewrow(new.entry))z <- do.call(rbind, lapply(c(new.entry, new.entry), addnewrow))
rbind(fruit.region, z)fruit.region[] <- lapply(fruit.region, as.character)