将大规模字符转换为日期格式,如r中的字符
我有一个1000万行的数据帧将大规模字符转换为日期格式,如r中的字符,r,string,stringi,R,String,Stringi,我有一个1000万行的数据帧df。我想将“生日”列的字符格式从“xxxxxxxx”转换为“XXXXXX”。例如,从“20051023”到“2005-10-23”。我可以使用df$birthdayGo和sub date <- c("20051023", "20151023") sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", date) # [1] "2005-10-23" "2015-10-23" dateas.date对向量有效 df$
df
。我想将“生日”列的字符格式从“xxxxxxxx”转换为“XXXXXX”。例如,从“20051023”到“2005-10-23”。我可以使用df$birthdayGo和sub
date <- c("20051023", "20151023")
sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", date)
# [1] "2005-10-23" "2015-10-23"
dateas.date
对向量有效
df$birthday <- format(as.Date(df$birthday, "%Y%m%d"), "%Y-%m-%d)
sub()
适用于矩阵,但不适用于data.frames。因此,as.matrix
df <- as.data.frame(matrix("20051023", ncol = 3, nrow = 3))
df$ID <- seq_len(nrow(df))
df[, 1:3] <- sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", as.matrix(df[, 1:3]))
哦,天哪。这真的很有帮助。谢谢你,Avinash Rai。as.character(as.Date(df$birth,“%Y%m%d”)
应该这样做这是真的。但是format()比as.character()快。见基准。谢谢,蒂埃里。如果我有两个或更多的列要转换,例如df[,c(3,5,7)],没有apply我怎么能转换呢?有没有什么方法能像.Date()或正则表达式那样快?对于(c(3,5,7)中的i){df[,我]谢谢你教我这项技能。这对我提高编码技能和概念非常有帮助。非常感谢你,蒂埃里。
microbenchmark(
as.character(as.Date(df$birthday, "%Y%m%d")),
format(as.Date(df$birthday, "%Y%m%d"), "%Y-%m%-d"),
sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", df$birthday)
)
Unit: microseconds
expr min lq mean
as.character(as.Date(df$birthday, "%Y%m%d")) 4923.189 5057.462 5390.313
format(as.Date(df$birthday, "%Y%m%d"), "%Y-%m%-d") 3428.657 3553.736 3697.660
sub("^(\\\\d{4})(\\\\d{2})(\\\\d{2})$", "\\\\1-\\\\2-\\\\3", df$birthday) 713.699 739.997 815.737
median uq max neval cld
5150.0420 5394.4265 8225.270 100 c
3594.7875 3665.9865 5753.200 100 b
763.0885 783.1865 2433.585 100 a
df <- as.data.frame(matrix("20051023", ncol = 3, nrow = 3))
df$ID <- seq_len(nrow(df))
df[, 1:3] <- sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", as.matrix(df[, 1:3]))
df <- as.data.frame(matrix("20051023", ncol = 20, nrow = 3))
df$ID <- seq_len(nrow(df))
library(microbenchmark)
microbenchmark(
matrix = df[, seq_len(ncol(df) - 1)] <- sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", as.matrix(df[, seq_len(ncol(df) - 1)])),
forloop = {
for(i in seq_len(ncol(df) - 1)){
df[, i] <- sub("^(\\d{4})(\\d{2})(\\d{2})$", "\\1-\\2-\\3", df[, i])
}
}
)
Unit: microseconds
expr min lq mean median uq max neval cld
matrix 460.555 476.805 504.3012 494.1235 507.594 1122.522 100 a
forloop 1554.425 1590.774 1677.3038 1625.8390 1670.312 3563.845 100 b