优雅地将1M转换为1000000 [英] Converting 1M to 1000000 elegantly
本文介绍了优雅地将1M转换为1000000的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我要转换:
library(data.table)
market.cap <- data.table(cap=c("1B", "10M", "2M"))
cap
1 1B
2 10M
3 2M
到:
cap
1 1000000000
2 10000000
3 2000000
我的解决方案。它工作,但涉及添加一列,我知道是不必要的。什么是更好的方法?
Here's my solution. It works, but involves adding a column, which I know isn't necessary. What's a better way?
market.cap[, cap1 := cap]
market.cap$cap = sapply(market.cap$cap, function(x) (as.numeric(temp <- gsub("B", "", x)) * 1000000000))
market.cap$cap1 = sapply(market.cap$cap1, function(x) (as.numeric(temp <- gsub("M", "", x)) * 1000000))
M = data.frame(x = na.omit(market.cap$cap))
B = data.frame(x = na.omit(market.cap$cap1))
rbind(M,B)
推荐答案
这是我自己的尝试:
market.cap[ , cap1 := {
sf <- gsub("[0-9]", "", cap)
as.numeric(gsub("[^0-9]", "", cap)) * 1000 ^ (2 + (sf == "B"))}]
以下方法可能更快,因为它不需要浪费努力通过正则表达式运行 cap
两次:
The following approach may prove faster since it doesn't need to waste effort on running cap
through a regex twice:
market.cap[ , cap1 := {
x<- do.call("rbind", strsplit(cap, split = "(?=[BM])", perl = TRUE))
as.numeric(x[ , 1L]) * 1000 ^ (2 + (x[ , 2L] == "B"))}]
并且以下可以证明是最快的,因为 tstrsplit
code> data.table :
And the following may prove fastest since tstrsplit
has been optimized in data.table
:
market.cap[ , cap1 := {
x <- tstrsplit(cap, split = "(?=[BM])", perl = TRUE)
as.numeric(x[[1L]]) * 1000 ^ (2 + (x[[2]] == "B"))}]
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