从维基百科导入表后,我有以下表格的值列表:
> tbl[2:6]
$`Internet
Explorer`
[1] "30.71%" "30.78%" "31.23%" "32.08%" "32.70%" "32.85%" "32.04%" "32.31%" "32.12%" "34.07%" "34.81%"
[12] "35.75%" "37.45%" "38.65%" "40.63%" "40.18%" "41.66%" "41.89%" "42.45%" "43.58%" "43.87%" "44.52%"
$Chrome
[1] "36.52%" "36.42%" "35.72%" "34.77%" "34.21%" "33.59%" "33.81%" "32.76%" "32.43%" "31.23%" "30.87%"
[12] "29.84%" "28.40%" "27.27%" "25.69%" "25.00%" "23.61%" "23.16%" "22.14%" "20.65%" "19.36%" "18.29%"
我试图摆脱百分号,以便将数据转换为数字形式。
是否有更快的方法来清理此数据而不是进行矢量化?我的当前代码如下:
data <- lapply(tbl[2:6], FUN = function(x) as.numeric(gsub("%", "", x)))
数据最终成为数据框,但我无法让gsub
在数据框的所有元素上正常工作。 有没有办法gsub()数据框的每个元素?
答案 0 :(得分:12)
我认为你可以通过以下方式实现,但我不知道它是否比你的更好或更清洁:
df <- data.frame(tbl)
df[,-1] <- as.numeric(gsub("%", "", as.matrix(df[,-1])))
给出了:
R> head(df)
Date Internet.Explorer Chrome Firefox Safari Opera Mobile
1 January 2013 30.71 36.52 21.42 8.29 1.19 14.13
2 December 2012 30.78 36.42 21.89 7.92 1.26 14.55
3 November 2012 31.23 35.72 22.37 7.83 1.39 13.08
4 October 2012 32.08 34.77 22.32 7.81 1.63 12.30
5 September 2012 32.70 34.21 22.40 7.70 1.61 12.03
6 August 2012 32.85 33.59 22.85 7.39 1.63 11.78
R> sapply(df, class)
Date Internet.Explorer Chrome Firefox
"factor" "numeric" "numeric" "numeric"
Safari Opera Mobile
"numeric" "numeric" "numeric"
答案 1 :(得分:4)
像朱巴一样,我不确定这种方式是“更好还是更干净”但是......要对数据框的所有元素采取行动,你可以使用 apply :
# start with data frame, not list
url <- "http://en.wikipedia.org/wiki/Usage_share_of_web_browsers"
# Get the eleventh table.
tbl <- readHTMLTable(url, which = 11, stringsAsFactors = F)
# use apply on the non-date columns
tbl[, 2:7] <- apply(tbl[, 2:7], 2, function(x) as.numeric(gsub("%", "", x)))