r 将字符串填充到相同的长度

经过几个小时的搜索应该很简单的内容后，我需要帮助。

我想做的事： 确保所有字符串都填充为相同的 26 个字符长度。

Dataset:

  library(stringr)

  names <-
  structure(list(
    names = c(
      "A",
      "ABC",
      "ABCDEFG",
      "ABCDEFGHIJKLMNOP",
      "AB",
      "ABCDEFGHI",
      "ABCDEFGHIJKLMNOPQRSTUVWXYZ",
      "ABCDEFGHIJKL",
      "ABCDEFGHIJKLMNOPQR",
      "ABCDEFGHIJKLMNOP",
      "ABCDEFGHIJKLMNO"
    )
  ),
  class = "data.frame",
  row.names = c(NA,-11L))

步骤1： 查找最大字符长度和要填充的空格数：

max <- as.numeric(max(nchar(names$names)))
max

n <- as.numeric(nchar(names$names))
n

pad <- max - n
pad


#add columns to the dataset to check how many characters are to be padded for each name

names$max <- as.numeric(max(nchar(names$names)))
names$n <- as.numeric(nchar(names$names))
names$pad <- as.numeric(max - n)

第 2 步：垫

  names$names <-
  str_pad(names$names,
          pad,
          side = "right",
          pad = "0")

但这种方法似乎对我不起作用。有人能指出我正确的方向吗？我得到不同长度的字符串：

                        names max  n pad
1   A000000000000000000000000  26  1  25
2     ABC00000000000000000000  26  3  23
3         ABCDEFG000000000000  26  7  19
4            ABCDEFGHIJKLMNOP  26 16  10
5    AB0000000000000000000000  26  2  24
6           ABCDEFGHI00000000  26  9  17
7  ABCDEFGHIJKLMNOPQRSTUVWXYZ  26 26   0
8              ABCDEFGHIJKL00  26 12  14
9          ABCDEFGHIJKLMNOPQR  26 18   8
10           ABCDEFGHIJKLMNOP  26 16  10
11            ABCDEFGHIJKLMNO  26 15  11

帮助将不胜感激。

这里我们只需要

library(dplyr)
mx <- as.numeric(max(nchar(names$Name)))
names$Name <- str_pad(names$Name, mx, side = "right", pad = "0")
names$Name

-output

#[1] "A0000000000000000000000000" "ABC00000000000000000000000" "ABCDEFG0000000000000000000" "ABCDEFGHIJKLMNOP0000000000"
#[5] "AB000000000000000000000000" "ABCDEFGHI00000000000000000" "ABCDEFGHIJKLMNOPQRSTUVWXYZ" "ABCDEFGHIJKL00000000000000"
#[9] "ABCDEFGHIJKLMNOPQR00000000" "ABCDEFGHIJKLMNOP0000000000" "ABCDEFGHIJKLMNO00000000000"

注意：最好不要使用函数名或参数名来命名对象