我有一个包含1000万行的数据框df
。我想将“birthday”列的字符格式从“xxxxxxxx”转换为“xxxx-xx-xx”。例如。从“20051023”到“2005-10-23”。我可以使用df$birthday <- lapply(df$birthday, as.Date, "%Y%m%d")
来做到这一点,但它浪费了大量的内存和计算时间进行数据转换。但是,我只想将其转换为类似日期的字符,而不是日期类型。因此我使用stringi
包,因为它是由C语言编写的。不幸的是,df$birthday <- stri_join(stri_sub(df$birthday, from=c(1,5,7), to=c(4,6,8)), collapse = "-")
不起作用,因为该函数不支持向量输入。有什么方法可以解决这个问题吗?非常感谢。
答案 0 :(得分:1)
as.Date
适用于矢量
df$birthday <- format(as.Date(df$birthday, "%Y%m%d"), "%Y-%m-%d)
矢量化函数比应用
快得多library(microbenchmark)
n <- 1e3
df <- data.frame(birthday = rep("20051023", n))
microbenchmark(
lapply(df$birthday, as.Date, "%Y%m%d"),
sapply(df$birthday, as.Date, "%Y%m%d"),
as.Date(df$birthday, "%Y%m%d")
)
Unit: microseconds
expr min lq mean median uq max neval cld
lapply(df$birthday, as.Date, "%Y%m%d") 22833.624 25340.118 29064.7188 28406.154 32346.245 58522.360 100 b
sapply(df$birthday, as.Date, "%Y%m%d") 24048.493 26252.660 29797.9074 28437.156 33119.381 47966.133 100 b
as.Date(df$birthday, "%Y%m%d") 431.469 447.719 481.5221 461.189 475.086 1984.158 100 a
正则表达式更偏快。
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
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]))
矩阵解决方案比for循环更快。差异随着您需要循环的列数而增加。
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