将前缀和后缀添加到列表值R

Question

我的R代码

data <- read.csv('filename.csv')
typof(data)
[1] "list"
str(data)
    'data frame' : 9 obs. of 10 variables
    $Name: Factor w/9 levels "Name 1", "Name 2",....
    $Column2: chr "","Text1","","Text2"
    $Column3: chr "Text2","Text3","","Text1"
    $Column4: chr "","","","Text1"
#and so on

要求：

我想要的只是在$Column2，$Column3，$Column4中，...在没有空白值的情况下，添加前缀（Here this is）和后缀（completed）。因此，考虑到data Column2上方，目前具有值"Text1"的第二行应该成为"Here this is Text1 completed."

类似地，在Column3: 1st, 2nd and 4th cell中需要添加前缀和后缀值。

i 不想使用loop ，除非并取消直接/necsary。

我的尝试：

我尝试了很少的尝试，例如 interaction， mget， append，而更多的尝试似乎没有用。

Answer 1

我会如下矢量化

indx <- which(data[, -1] != "", arr.ind = TRUE) # Find all non-empty incidences 
data[, -1][indx] <- paste("Here this is", data[, -1][indx], "completed.")

Answer 2

这将适用于前4列

apply(data[,2:4],2,function(x) ifelse(x != "",paste("Here this is ",x," completed."),x))

假设每个列的前缀和后缀相同。它确实返回了矩阵，但是将其转换为数据框架很容易。希望它有帮助。

编辑：刚刚意识到您的数据在列表中，因此您需要lapply或sapply。类似：

sapply(data,function(x) ifelse(x != "",paste("Here this is ",x," completed."),x))[,2:4]

还返回一个矩阵。

Answer 3

以下是set的选项，该选项在不复制的情况下分配到位

library(data.table)
setDT(data)
for(j in 2:ncol(data)){
      set(data, i = which(data[[j]]!=""), 
                j = j,
                value = paste("Here there is ",  data[[j]][data[[j]]!=""], " completed."))
    }
data
#     Name                          Column2                          Column3
#1: Name 1                                  Here there is  Text1  completed.
#2: Name 2 Here there is  Text1  completed.                                 
#3: Name 3 Here there is  Text2  completed. Here there is  Text2  completed.
#4: Name 4                                  Here there is  Text3  completed.

数据

data <- structure(list(Name = structure(1:4, .Label = c("Name 1", "Name 2", 
"Name 3", "Name 4"), class = "factor"), Column2 = c("", "Text1", 
 "Text2", ""), Column3 = c("Text1", "", "Text2", "Text3")), .Names = c("Name", 
 "Column2", "Column3"), row.names = c(NA, -4L), class = "data.frame")

Answer 4

使用 lapply函数（也是循环）：

# dummy data
df1 <- mtcars[1:5, 1:3]
# add blanks
df1[2,2] <- ""
df1
#                    mpg cyl disp
# Mazda RX4         21.0   6  160
# Mazda RX4 Wag     21.0      160
# Datsun 710        22.8   4  108
# Hornet 4 Drive    21.4   6  258
# Hornet Sportabout 18.7   8  360
# add prefix and suffix
res <- cbind(df1[, 1, drop = FALSE],
             data.frame(
               lapply(df1[, -1], function(i)
                 ifelse(i == "", i, paste("Here this is", i, "completed.")))))

res
#                    mpg                       cyl                        disp
# Mazda RX4         21.0 Here this is 6 completed. Here this is 160 completed.
# Mazda RX4 Wag     21.0                           Here this is 160 completed.
# Datsun 710        22.8 Here this is 4 completed. Here this is 108 completed.
# Hornet 4 Drive    21.4 Here this is 6 completed. Here this is 258 completed.
# Hornet Sportabout 18.7 Here this is 8 completed. Here this is 360 completed.

Answer 5

DF = transform(ifelse(data$Column2 == "", data$Column2, sprintf('Here it is %s completed', data$Column2)))
DF <- data.frame (DF, data$Name, data$Column3)