Question

我有一个像这样的数据框：

  id subscriberid  intName
1  1   1234567890 asdfsadf
2  2   3243245324  dfsafdf
3  3   4532453245  dasdfsd

可重现的示例如下：

structure(list(id = 1:3, subscriberid = c(1234567890, 3243245324, 4532453245),
    intName = c("asdfsadf", "dfsafdf", "dasdfsd")),
    row.names = c(NA, 3L), class = "data.frame")

我有一个subid数组，只要subid与df $ subscriberid匹配，我都必须将订户ID的第一位更改为9。

subid = c(1234567890,2345345234)

我尝试了以下操作：

for (i in df$subscriberid) {
    df$subscriberid == sub(substr(df$subscriberid,0,1),9,df$subscriberid)
}

我也尝试过将ifelse与substr和gsub以及其他不同的组合一起使用。但是无法通过。所需的输出是

  id subscriberid  intName
1  1   9234567890 asdfsadf   <--- only the first digit is changed.
2  2   3243245324  dfsafdf
3  3   4532453245  dasdfsd

Answer 1

一种选择是使用ifelse，如果subscriberid中存在subid，那么我们paste 9的剩余字符串将从第二个索引开始。

df$subscriberid <- with(df, ifelse(subscriberid %in% subid,
                    paste0("9",substring(subscriberid,2)), subscriberid))

df
#  id subscriberid  intName
#1  1   9234567890 asdfsadf
#2  2   3243245324  dfsafdf
#3  3   4532453245  dasdfsd

使用substring的好处是您只需要提及起始索引（此处为2），stop的默认值为1000000，覆盖了大多数字符串。

Answer 2

我们可以尝试构建匹配ID的正则表达式模式，然后使用grepl在数据框中找到匹配的行：

regex <- paste0("\\b(", paste(subid, collapse="|"), ")\\b")
df$subscriberid <- ifelse(grepl(regex, df$subscriberid),
    paste0("9", substr(df$subscriberid, 2, nchar(df$subscriberid))),
    df$subscriberid)

df
  id subscriberid  intName
1  1   9234567890 asdfsadf
2  2   3243245324  dfsafdf
3  3   4532453245  dasdfsd

Answer 3

一种使用数学的方法，其好处是将结果以数字形式返回，并且速度更快。

数据

df <- structure(list(id = 1:3, subscriberid = c(1234567890, 3243245324, 4532453245),
                     intName = c("asdfsadf", "dfsafdf", "dasdfsd")),
                row.names = c(NA, 3L), class = "data.frame")

subid <- c(1234567890,2345345234)

方法

idx <- df$subscriberid %in% subid
vals <- df[ idx, "subscriberid" ]
digits <- floor( log10( vals ) )

## number of digits given by `floor( log10( vals) ) + 1`, but we want the first digit

( ( vals / 10^digits ) + 9 - floor( vals / 10^digits ) ) * (10^digits)

# [1] 9234567890

这是在寻找与subid

相匹配的data.frame索引

然后是

使用log10来计算这些数字中有多少位数
将这些数字的幂除以10，然后对其进行累加以获得第一个整数
从9（您的目标）中减去该整数
将其添加回log10的值
将这些数字的幂乘以10，以获得原始数字位数

基准

library(microbenchmark)

microbenchmark(
  ronak = { ronak( df, subid ) },
  tim = { tim( df, subid ) },
  tmfmnk = { tmfmnk( df, subid ) },
  symbolix = { symbolix( df, subid ) },
  times = 5
)

# Unit: milliseconds
# expr            min         lq       mean     median         uq      max neval
# ronak    186.143804 188.618750 214.151592 191.154106 196.399341 308.4420     5
# tim      442.385985 463.510154 526.814255 506.268620 541.829769 680.0767     5
# tmfmnk   236.423472 255.418334 295.652617 295.624544 329.901976 360.8948     5
# symbolix   5.510366   5.828804   8.166222   5.850937   5.942607  17.6984     5

并显示结果相等

res_ronak <- ronak( df, subid )
res_tim <- tim( df, subid )
res_tmfmnk <- tmfmnk( df, subid )
res_symbolix <- symbolix( df, subid )

all.equal(res_ronak, res_tim)
# [1] TRUE
all.equal(res_tim, res_tmfmnk)
# [1] TRUE
res_symbolix$subscriberid <- as.character(res_symbolix$subscriberid)
all.equal(res_tmfmnk, res_symbolix)
# [1] TRUE

基准化数据

set.seed(1234)
df <- data.frame(
  subscriberid = sample(1:100000000, size = 1e5)
)
subid <- sample( df$subscriberid, size = 10 )

基准化功能

ronak <- function(df, subid) {
  df$subscriberid <- with(df, ifelse(subscriberid %in% subid,
                  paste0("9",substring(subscriberid,2)), subscriberid))
  return(df)
}

tim <- function(df, subid) {
  regex <- paste0("\\b(", paste(subid, collapse="|"), ")\\b")
  df$subscriberid <- ifelse(grepl(regex, df$subscriberid),
                            paste0("9", substr(df$subscriberid, 2, nchar(df$subscriberid))),
                            df$subscriberid)
  return(df)
}

tmfmnk <- function(df, subid) {
  df$subscriberid <- ifelse(df$subscriberid %in% subid, 
         sub(".", "9", df$subscriberid), df$subscriberid)
  return(df)
}

symbolix <- function(df, subid) {
  idx <- df$subscriberid %in% subid
  vals <- df[ idx, "subscriberid" ]
  digits <- floor( log10( vals ) )
  df[ idx, "subscriberid" ] <- ( ( vals / 10^digits ) + 9 - floor( vals / 10^digits ) ) * (10^digits)
  return(df)
}

Answer 4

使用sub()的可能性稍有不同：

df$subscriberid <- ifelse(df$subscriberid %in% subid, 
                   sub(".", "9", df$subscriberid), df$subscriberid)

  id subscriberid  intName
1  1   9234567890 asdfsadf
2  2   3243245324  dfsafdf
3  3   4532453245  dasdfsd

如果“ subscriberid”与“ subid”匹配，则“ subscriberid”中的第一个字符将替换为9，否则保持不变。

Answer 5

更容易使用substring的分配方法

# create a logical vector
i1 <- df1$subscriberid %in% subid

# convert the column to character class
df1$subscriberid <- as.character(df1$subscriberid)

#assign with substring<-
substring(df1$subscriberid[i1], 1, 1) <- '9'
df1
#   id subscriberid  intName
#1  1   9234567890 asdfsadf
#2  2   3243245324  dfsafdf
#3  3   4532453245  dasdfsd

基准

包括此方法以及@SymbolixAU的数据（其他功能取自他的帖子）

akrun <- function(df, subid) {
 i1 <- df$subscriberid %in% subid
 df$subscriberid <- as.character(df$subscriberid)
 substring(df$subscriberid[i1], 1, 1) <- '9'
  }


set.seed(1234)
df <- data.frame(
   subscriberid = sample(1:100000000, size = 1e5)
  )
subid <- sample( df$subscriberid, size = 10 )
library(microbenchmark)
microbenchmark(
   ronak = { ronak( df, subid ) },
   tim = { tim( df, subid ) },
   tmfmnk = { tmfmnk( df, subid ) },
   symbolix = { symbolix( df, subid ) }, akrun = {akrun(df, subid)}, times = 5)
#Unit: milliseconds
#     expr        min         lq       mean     median        uq       max neval cld
#    ronak 105.073716 128.279151 140.993520 138.241632 154.89092 178.48218     5  b 
#      tim 224.610660 246.959505 263.138679 264.685503 284.93632 294.50141     5   c
#   tmfmnk 119.734979 134.949406 138.735054 135.888113 142.91750 160.18527     5  b 
# symbolix   2.487283   3.238862   8.429718   3.540119  10.80669  22.07564     5 a  
#    akrun  29.530330  33.431953  41.649046  34.772512  36.91314  73.59730     5 a

数据

df1 <- structure(list(id = 1:3, subscriberid = c(1234567890, 3243245324, 4532453245),
intName = c("asdfsadf", "dfsafdf", "dasdfsd")),
row.names = c(NA, 3L), class = "data.frame")

根据条件更改列中的第一个数字

5 个答案:

数据

方法

基准

基准化数据

基准化功能

基准

数据