使用 R 将列名插入其值
Insert Column Name into its Value using R
我需要在其值中插入列名称、部门。我有这样的代码:
Department <- c("Store1","Store2","Store3","Store4","Store5")
Department2 <- c("IT1","IT2","IT3","IT4","IT5")
x <- c(100,200,300,400,500)
Result <- data.frame(Department,Department2,x)
Result
预期结果如下:
Department <- c("Department_Store1","Departmentz_Store2","Department_Store3","Department_Store4","Department_Store5")
Department2 <- c("Department2_IT1","Department2_IT2","Department2_IT3","Department2_IT4","Department2_IT5")
x <- c(100,200,300,400,500)
Expected.Result <- data.frame(Department,Department2,x)
Expected.Result
有人可以帮忙吗?谢谢
dplyr
和 tidyr
的另一种方式:
library(dplyr)
library(tidyr)
# Convert to character to avoid warning message, will convert all columns to character
Result[] <- lapply(Result, as.character)
Result %>%
mutate_if(is.factor, as.character) %>% # optional, only convert factor to character, retain all other types
gather(key, value, -x) %>%
mutate(var = paste(key, value, sep = "_")) %>%
select(-value) %>%
spread(key,var)
x Department Department2
1 100 Department_Store1 Department2_IT1
2 200 Department_Store2 Department2_IT2
3 300 Department_Store3 Department2_IT3
4 400 Department_Store4 Department2_IT4
5 500 Department_Store5 Department2_IT5
数据:
Result <- data.frame(
Department = c("Store1","Store2","Store3","Store4","Store5"),
Department2 = c("IT1","IT2","IT3","IT4","IT5"),
x = c(100,200,300,400,500)
)
如果您将有问题的列名收集到向量中 dep_col
,这是一个带有 for 循环的干净 base R
解决方案:
df <- data.frame(x = 1:5,
Department = paste0("Store", 1:5),
Department2 = paste0("IT", 1:5))
dep_col <- names(df)[-1]
for (c in dep_col)
df[[c]] <- paste(c, df[[c]], sep = "_")
如果我理解正确,OP 希望在所有以 "Department"
开头的列中的值前面加上相应的列名。
编辑 根据 OP 的要求,select 列的代码已被概括为选择其他列名称。
这是一个使用 data.table
的快速 set()
函数的解决方案:
library(data.table)
setDT(Result)
cols <- stringr::str_subset(names(Result), "^(Department|Division|Team)")
for (j in cols) {
set(Result, NULL, j, paste(j, Result[[j]], sep = "_"))
}
Result
Department Department2 x
1: Department_Store1 Department2_IT1 100
2: Department_Store2 Department2_IT2 200
3: Department_Store3 Department2_IT3 300
4: Department_Store4 Department2_IT4 400
5: Department_Store5 Department2_IT5 500
请注意 set()
通过引用更新,即不复制整个对象。
我需要在其值中插入列名称、部门。我有这样的代码:
Department <- c("Store1","Store2","Store3","Store4","Store5")
Department2 <- c("IT1","IT2","IT3","IT4","IT5")
x <- c(100,200,300,400,500)
Result <- data.frame(Department,Department2,x)
Result
预期结果如下:
Department <- c("Department_Store1","Departmentz_Store2","Department_Store3","Department_Store4","Department_Store5")
Department2 <- c("Department2_IT1","Department2_IT2","Department2_IT3","Department2_IT4","Department2_IT5")
x <- c(100,200,300,400,500)
Expected.Result <- data.frame(Department,Department2,x)
Expected.Result
有人可以帮忙吗?谢谢
dplyr
和 tidyr
的另一种方式:
library(dplyr)
library(tidyr)
# Convert to character to avoid warning message, will convert all columns to character
Result[] <- lapply(Result, as.character)
Result %>%
mutate_if(is.factor, as.character) %>% # optional, only convert factor to character, retain all other types
gather(key, value, -x) %>%
mutate(var = paste(key, value, sep = "_")) %>%
select(-value) %>%
spread(key,var)
x Department Department2
1 100 Department_Store1 Department2_IT1
2 200 Department_Store2 Department2_IT2
3 300 Department_Store3 Department2_IT3
4 400 Department_Store4 Department2_IT4
5 500 Department_Store5 Department2_IT5
数据:
Result <- data.frame(
Department = c("Store1","Store2","Store3","Store4","Store5"),
Department2 = c("IT1","IT2","IT3","IT4","IT5"),
x = c(100,200,300,400,500)
)
如果您将有问题的列名收集到向量中 dep_col
,这是一个带有 for 循环的干净 base R
解决方案:
df <- data.frame(x = 1:5,
Department = paste0("Store", 1:5),
Department2 = paste0("IT", 1:5))
dep_col <- names(df)[-1]
for (c in dep_col)
df[[c]] <- paste(c, df[[c]], sep = "_")
如果我理解正确,OP 希望在所有以 "Department"
开头的列中的值前面加上相应的列名。
编辑 根据 OP 的要求,select 列的代码已被概括为选择其他列名称。
这是一个使用 data.table
的快速 set()
函数的解决方案:
library(data.table)
setDT(Result)
cols <- stringr::str_subset(names(Result), "^(Department|Division|Team)")
for (j in cols) {
set(Result, NULL, j, paste(j, Result[[j]], sep = "_"))
}
Result
Department Department2 x 1: Department_Store1 Department2_IT1 100 2: Department_Store2 Department2_IT2 200 3: Department_Store3 Department2_IT3 300 4: Department_Store4 Department2_IT4 400 5: Department_Store5 Department2_IT5 500
请注意 set()
通过引用更新,即不复制整个对象。