从 R 中的多个列中删除字符串字符并进行变异

Question

这是我的数据框（tables_df）的头部，都是字符类型：

#        Name Ticker  Last price       %     %7d    %30d    %90d    %180d    %365d # Coins
1 1     bitcoin    BTC $ 56,667.17   1.33%  21.13%  74.43% 209.62%  386.02%  489.33%  18.63M
2 2    ethereum    ETH $ 1,976.941   1.00%  10.37%  60.49% 269.53%  404.93%  662.55% 114.75M
3 3 binancecoin    BNB $ 265.20462 -20.41% 102.40% 553.29% 791.32% 1114.70% 1083.61%  99.01M
4 4     cardano    ADA $ 1.1537436  24.50%  31.18% 239.85% 765.91%  849.66% 1909.86%  25.92B
5 5    litecoin    LTC $ 231.95832  -2.28%   4.65%  72.05% 180.01%  282.54%  227.14%  66.08M
6 6     uniswap    UNI $ 30.304469  50.48%  41.73% 268.43% 732.38%       0%       0% 126.20M
  Market cap MC rank
1    $ 1.05T       1
2  $ 226.86B       2
3   $ 26.25B       5
4   $ 29.91B       4
5   $ 15.32B       7
6    $ 3.82B      23

我想在 R 中做 3 件事

1. 将列名中包含“%”的所有列中的百分比字符转换为数字百分比值
2. 将具有 M（百万）或 B（十亿）或 T（万亿）的字符转换为“Marcet cap”和“Coins”列中的数值（x1000.000 表示百万，x1.000.000.000 表示十亿和 x1。 000.000.000.000 表示万亿）
3. 删除美元符号并将“最后价格”和“市值”列中的数字转换为数字

第一件事我尝试了这段代码：

tables_df %>% mutate_at(vars(contains('%')), str_remove(string = .,pattern = "%")) %>%
  mutate_at(vars(contains('%')),funs =  as.numeric())

还有这段代码：

tables_df %>% parse_number("%", trim_ws = TRUE)

很遗憾，这两种尝试都不起作用。

对于我的第二和第三件事，我还没想好从哪里开始。有人可以帮我吗？非常感谢！

dput(head(tables_df))
structure(list(`#` = c("1", "2", "3", "4", "5", "6"), c("", "", 
"", "", "", ""), Name = c("bitcoin", "ethereum", "binancecoin", 
"cardano", "litecoin", "uniswap"), Ticker = c("BTC", "ETH", "BNB", 
"ADA", "LTC", "UNI"), `Last price` = c("$ 57,431.62", "$ 1,959.472", 
"$ 302.91449", "$ 1.1072817", "$ 230.71046", "$ 29.981031"), 
    `%` = c("2.59%", "2.21%", "19.01%", "-0.96%", "1.50%", "4.01%"
    ), `%7d` = c("17.24%", "7.72%", "123.90%", "32.27%", "5.13%", 
    "38.44%"), `%30d` = c("80.95%", "63.35%", "648.16%", "234.17%", 
    "66.21%", "259.22%"), `%90d` = c("211.37%", "236.87%", "893.12%", 
    "617.17%", "165.07%", "688.65%"), `%180d` = c("388.06%", 
    "379.98%", "1238.50%", "793.63%", "271.27%", "0%"), `%365d` = c("496.72%", 
    "658.28%", "1263.44%", "1798.47%", "231.57%", "0%"), `# Coins` = c("18.63M", 
    "114.76M", "99.01M", "25.92B", "66.08M", "126.20M"), `Market cap` = c("$ 1.07T", 
    "$ 224.87B", "$ 29.99B", "$ 28.70B", "$ 15.24B", "$ 3.78B"
    ), `MC rank` = c("1", "2", "4", "5", "7", "23")), row.names = c(NA, 
6L), class = "data.frame")

Answer 1

编写一个函数将百万/十亿/万亿更改为实际数字。

change_num <- function(x) {
  x1 <- parse_number(x)
  x1 * case_when(grepl('T', x) ~ 1e12, 
                 grepl('B', x) ~ 1e9, 
                 grepl('M', x) ~ 1e6)
}

使用across 将函数应用于多个列。

library(dplyr)
library(readr)

tables_df <- tables_df[-2] %>%
                mutate(across(`Last price`:`%365d`, parse_number),
                  across(c(`# Coins`,`Market cap`), change_num))

tables_df

#           Name Ticker   Last price     %    %7d   %30d   %90d   %180d   %365d
#1 1     bitcoin    BTC 57431.620000  2.59  17.24  80.95 211.37  388.06  496.72
#2 2    ethereum    ETH  1959.472000  2.21   7.72  63.35 236.87  379.98  658.28
#3 3 binancecoin    BNB   302.914490 19.01 123.90 648.16 893.12 1238.50 1263.44
#4 4     cardano    ADA     1.107282 -0.96  32.27 234.17 617.17  793.63 1798.47
#5 5    litecoin    LTC   230.710460  1.50   5.13  66.21 165.07  271.27  231.57
#6 6     uniswap    UNI    29.981031  4.01  38.44 259.22 688.65    0.00    0.00

#     # Coins Market cap MC rank
#1 1.8630e+07 1.0700e+12       1
#2 1.1476e+08 2.2487e+11       2
#3 9.9010e+07 2.9990e+10       4
#4 2.5920e+10 2.8700e+10       5
#5 6.6080e+07 1.5240e+10       7
#6 1.2620e+08 3.7800e+09      23