在 R 中，将 "Oct 29 - Nov 1" 转换为“20201029”和“20201101”
In R, convert "Oct 29 - Nov 1" to "20201029" and "20201101"
我正在处理从网站上抓取的凌乱 table，为了使日期栏更有用，我需要清理抓取的内容。我们的数据看起来像这样：
mydata <- structure(list(Dates = c("Sep\r\n \r\n 10 - 13",
"Oct\r\n \r\n 8 - 11", "Oct 29 - Nov 1",
"Nov\r\n \r\n 19 - 22", "Jan\r\n \r\n 21 - 24",
"Mar\r\n \r\n 4 - 7", "Apr 29 - May 2"),
points = c("500", "500", "500", "500", "500", "550", "500"
)), row.names = c(1L, 5L, 8L, 11L, 16L, 23L, 32L), class = "data.frame")
> mydata
Dates points
1 Sep\r\n \r\n 10 - 13 500
5 Oct\r\n \r\n 8 - 11 500
8 Oct 29 - Nov 1 500
11 Nov\r\n \r\n 19 - 22 500
16 Jan\r\n \r\n 21 - 24 500
23 Mar\r\n \r\n 4 - 7 550
32 Apr 29 - May 2 500
Dates 中的每个日期都是一个日期范围，实际上应该是 startDate 和 endDate。那么我们要创建的是：
> newdata
StartDate EndDate points
1 20200910 20200913 500
1 20201008 20201011 500
1 20201029 20201101 500
1 20201119 20201122 500
1 20210121 20210124 500
1 20210304 20210307 500
1 20210429 20210502 500
我们可以假设 9 月 - 12 月的所有日期都是 2020 年，1 月 - 8 月的所有日期都是 2021 年。
编辑 1
这可能不是最简洁的代码，但我成功地将 Dates 列拆分为 2 列
cleaning_dates_df <- do.call('rbind', strsplit(mydata$Dates, '-')) %>% as.data.frame()
colnames(cleaning_dates_df) <- c('start', 'end')
cleaning_dates_df$start <- as.character(cleaning_dates_df$start)
cleaning_dates_df$end <- as.character(cleaning_dates_df$end)
cleaning_dates_df <- cleaning_dates_df %>%
dplyr::mutate(end = ifelse(nchar(end) > 4, end, paste0(trimws(sub("\r\n.*", "", start)), end))) %>%
dplyr::mutate(start = ifelse(nchar(start) < 8, start, paste0(trimws(sub("\r\n.*", "", start)), sub(".*\s", "", start)))) %>%
dplyr::mutate(end = trimws(end)) %>% dplyr::mutate(start = trimws(start))
head(cleaning_dates_df, 8)
...还需要转换成YYYYMMDD
我不会说它很漂亮，但是你可以先使用正则表达式来提取所有部分：
rgx <- "^([a-z]+)(\r|\n|\s)+(\d+)\s\-\s([a-z]+)*\s*(\d+)$"
td <- strcapture(rgx, tolower(mydata$Dates),
proto=list(mth1="",x="",day1="",mth2="",day2=""))
只提到一个月时重复月份：
td$mth2[td$mth2 == ''] <- td$mth1[td$mth2 == '']
将月份转换为数字，然后决定是 2020 年还是 2021 年：
td[c("mth1","mth2")] <- lapply(td[c("mth1","mth2")],
function(x) match(x, tolower(month.abb)))
td[c("yr1","yr2")] <- lapply(td[c("mth1","mth2")],
function(x) ifelse(x >= 9, 2020, 2021) )
从不同的部分构建日期：
mydata$startdate <- as.Date(paste(td$yr1, td$mth1, td$day1, sep="/"))
mydata$enddate <- as.Date(paste(td$yr2, td$mth2, td$day2, sep="/"))
完成！：
mydata
# Dates points startdate enddate
#1 Sep\r\n \r\n 10 - 13 500 2020-09-10 2020-09-13
#5 Oct\r\n \r\n 8 - 11 500 2020-10-08 2020-10-11
#8 Oct 29 - Nov 1 500 2020-10-29 2020-11-01
#11 Nov\r\n \r\n 19 - 22 500 2020-11-19 2020-11-22
#16 Jan\r\n \r\n 21 - 24 500 2021-01-21 2021-01-24
#23 Mar\r\n \r\n 4 - 7 550 2021-03-04 2021-03-07
#32 Apr 29 - May 2 500 2021-04-29 2021-05-02
你可以试试：
library(lubridate)
library(dplyr)
d <- do.call(rbind, lapply(str_split(gsub("[\v-]", " ", mydata$Dates), "\s+"), function(x) if (length(x) == 3) append(x, x[1], after = 2) else x) )
start_date <- as.Date(paste(d[,1], d[,2], "2020", sep = "-"), format = "%b-%d-%Y")
end_date <- as.Date(paste(d[,3], d[,4], "2020", sep = "-"), format = "%b-%d-%Y")
start_date <- if_else(month(start_date) < 9, start_date + years(1), start_date)
end_date <- if_else(month(end_date) < 9, end_date + years(1), end_date)
data.frame(start_date, end_date,mydata$points)
start_date end_date mydata.points
1 2020-09-10 2020-09-13 500
2 2020-10-08 2020-10-11 500
3 2020-10-29 2020-11-01 500
4 2020-11-19 2020-11-22 500
5 2021-01-21 2021-01-24 500
6 2021-03-04 2021-03-07 550
7 2021-04-29 2021-05-02 500
除非您有理由不这样做，否则最好以日期格式保存数据。但如果你需要它们作为显示的字符串，你可以使用 format()，例如：
format(start_date, "%Y%m%d")
这是一个凌乱的 Base R 解决方案：
# Function to pad dateparts: pad_dateparts => function()
pad_dateparts <- function(date_vec){
return(ifelse(nchar(date_vec) == 1, paste0("0", date_vec), date_vec))
}
# Store the months for each obersvation: months_ => list of characters
months_ <-
lapply(regmatches(mydata$Dates, gregexpr(
paste0(month.abb, collapse = "|"), mydata$Dates)), function(x) {
if (length(x) == 1) {
pad_dateparts(match(rep(x, 2), month.abb))
} else{
pad_dateparts(match(x, month.abb))
}
}
)
# Store the day numbers for each obersvation: days_ => list of characters
days_ <- lapply(sapply(trimws(gsub('\D+',' ', mydata$Dates), "both"), strsplit, "\s+"),
pad_dateparts)
# Function to increment years from ordered vector of month parts:
# increment_years => function()
increment_years <- function(start_year, ordered_month_vec){
return(start_year + cumsum(c(FALSE, diff(as.integer(ordered_month_vec)) < 0)))
}
# Store the year parts: years_ => list of data.frames
years_ <- split(apply(data.frame(do.call(rbind, months_)), 2,
function(x){increment_years(2020, x)}), seq_along(months_))
# Create the required data.frame: clean_df => data.frame
clean_df <- cbind(setNames(
data.frame(
do.call(rbind, Map(function(x, y, z) {
as.integer(paste0(x, y, z))
},
years_, months_, days_)),
row.names = NULL,
stringsAsFactors = FALSE
),
c("StartDate", "EndDate")
),
Points = mydata$points)
数据：
mydata <- structure(list(Dates = c("Sep\r\n \r\n 10 - 13",
"Oct\r\n \r\n 8 - 11", "Oct 29 - Nov 1",
"Nov\r\n \r\n 19 - 22", "Jan\r\n \r\n 21 - 24",
"Mar\r\n \r\n 4 - 7", "Apr 29 - May 2"),
points = c("500", "500", "500", "500", "500", "550", "500"
)), row.names = c(1L, 5L, 8L, 11L, 16L, 23L, 32L), class = "data.frame")
我正在处理从网站上抓取的凌乱 table，为了使日期栏更有用，我需要清理抓取的内容。我们的数据看起来像这样：
mydata <- structure(list(Dates = c("Sep\r\n \r\n 10 - 13",
"Oct\r\n \r\n 8 - 11", "Oct 29 - Nov 1",
"Nov\r\n \r\n 19 - 22", "Jan\r\n \r\n 21 - 24",
"Mar\r\n \r\n 4 - 7", "Apr 29 - May 2"),
points = c("500", "500", "500", "500", "500", "550", "500"
)), row.names = c(1L, 5L, 8L, 11L, 16L, 23L, 32L), class = "data.frame")
> mydata
Dates points
1 Sep\r\n \r\n 10 - 13 500
5 Oct\r\n \r\n 8 - 11 500
8 Oct 29 - Nov 1 500
11 Nov\r\n \r\n 19 - 22 500
16 Jan\r\n \r\n 21 - 24 500
23 Mar\r\n \r\n 4 - 7 550
32 Apr 29 - May 2 500
Dates 中的每个日期都是一个日期范围，实际上应该是 startDate 和 endDate。那么我们要创建的是：
> newdata
StartDate EndDate points
1 20200910 20200913 500
1 20201008 20201011 500
1 20201029 20201101 500
1 20201119 20201122 500
1 20210121 20210124 500
1 20210304 20210307 500
1 20210429 20210502 500
我们可以假设 9 月 - 12 月的所有日期都是 2020 年，1 月 - 8 月的所有日期都是 2021 年。
编辑 1
这可能不是最简洁的代码，但我成功地将 Dates 列拆分为 2 列
cleaning_dates_df <- do.call('rbind', strsplit(mydata$Dates, '-')) %>% as.data.frame()
colnames(cleaning_dates_df) <- c('start', 'end')
cleaning_dates_df$start <- as.character(cleaning_dates_df$start)
cleaning_dates_df$end <- as.character(cleaning_dates_df$end)
cleaning_dates_df <- cleaning_dates_df %>%
dplyr::mutate(end = ifelse(nchar(end) > 4, end, paste0(trimws(sub("\r\n.*", "", start)), end))) %>%
dplyr::mutate(start = ifelse(nchar(start) < 8, start, paste0(trimws(sub("\r\n.*", "", start)), sub(".*\s", "", start)))) %>%
dplyr::mutate(end = trimws(end)) %>% dplyr::mutate(start = trimws(start))
head(cleaning_dates_df, 8)
...还需要转换成YYYYMMDD
我不会说它很漂亮，但是你可以先使用正则表达式来提取所有部分：
rgx <- "^([a-z]+)(\r|\n|\s)+(\d+)\s\-\s([a-z]+)*\s*(\d+)$"
td <- strcapture(rgx, tolower(mydata$Dates),
proto=list(mth1="",x="",day1="",mth2="",day2=""))
只提到一个月时重复月份：
td$mth2[td$mth2 == ''] <- td$mth1[td$mth2 == '']
将月份转换为数字，然后决定是 2020 年还是 2021 年：
td[c("mth1","mth2")] <- lapply(td[c("mth1","mth2")],
function(x) match(x, tolower(month.abb)))
td[c("yr1","yr2")] <- lapply(td[c("mth1","mth2")],
function(x) ifelse(x >= 9, 2020, 2021) )
从不同的部分构建日期：
mydata$startdate <- as.Date(paste(td$yr1, td$mth1, td$day1, sep="/"))
mydata$enddate <- as.Date(paste(td$yr2, td$mth2, td$day2, sep="/"))
完成！：
mydata
# Dates points startdate enddate
#1 Sep\r\n \r\n 10 - 13 500 2020-09-10 2020-09-13
#5 Oct\r\n \r\n 8 - 11 500 2020-10-08 2020-10-11
#8 Oct 29 - Nov 1 500 2020-10-29 2020-11-01
#11 Nov\r\n \r\n 19 - 22 500 2020-11-19 2020-11-22
#16 Jan\r\n \r\n 21 - 24 500 2021-01-21 2021-01-24
#23 Mar\r\n \r\n 4 - 7 550 2021-03-04 2021-03-07
#32 Apr 29 - May 2 500 2021-04-29 2021-05-02
你可以试试：
library(lubridate)
library(dplyr)
d <- do.call(rbind, lapply(str_split(gsub("[\v-]", " ", mydata$Dates), "\s+"), function(x) if (length(x) == 3) append(x, x[1], after = 2) else x) )
start_date <- as.Date(paste(d[,1], d[,2], "2020", sep = "-"), format = "%b-%d-%Y")
end_date <- as.Date(paste(d[,3], d[,4], "2020", sep = "-"), format = "%b-%d-%Y")
start_date <- if_else(month(start_date) < 9, start_date + years(1), start_date)
end_date <- if_else(month(end_date) < 9, end_date + years(1), end_date)
data.frame(start_date, end_date,mydata$points)
start_date end_date mydata.points
1 2020-09-10 2020-09-13 500
2 2020-10-08 2020-10-11 500
3 2020-10-29 2020-11-01 500
4 2020-11-19 2020-11-22 500
5 2021-01-21 2021-01-24 500
6 2021-03-04 2021-03-07 550
7 2021-04-29 2021-05-02 500
除非您有理由不这样做，否则最好以日期格式保存数据。但如果你需要它们作为显示的字符串，你可以使用 format()，例如：
format(start_date, "%Y%m%d")
这是一个凌乱的 Base R 解决方案：
# Function to pad dateparts: pad_dateparts => function()
pad_dateparts <- function(date_vec){
return(ifelse(nchar(date_vec) == 1, paste0("0", date_vec), date_vec))
}
# Store the months for each obersvation: months_ => list of characters
months_ <-
lapply(regmatches(mydata$Dates, gregexpr(
paste0(month.abb, collapse = "|"), mydata$Dates)), function(x) {
if (length(x) == 1) {
pad_dateparts(match(rep(x, 2), month.abb))
} else{
pad_dateparts(match(x, month.abb))
}
}
)
# Store the day numbers for each obersvation: days_ => list of characters
days_ <- lapply(sapply(trimws(gsub('\D+',' ', mydata$Dates), "both"), strsplit, "\s+"),
pad_dateparts)
# Function to increment years from ordered vector of month parts:
# increment_years => function()
increment_years <- function(start_year, ordered_month_vec){
return(start_year + cumsum(c(FALSE, diff(as.integer(ordered_month_vec)) < 0)))
}
# Store the year parts: years_ => list of data.frames
years_ <- split(apply(data.frame(do.call(rbind, months_)), 2,
function(x){increment_years(2020, x)}), seq_along(months_))
# Create the required data.frame: clean_df => data.frame
clean_df <- cbind(setNames(
data.frame(
do.call(rbind, Map(function(x, y, z) {
as.integer(paste0(x, y, z))
},
years_, months_, days_)),
row.names = NULL,
stringsAsFactors = FALSE
),
c("StartDate", "EndDate")
),
Points = mydata$points)
数据：
mydata <- structure(list(Dates = c("Sep\r\n \r\n 10 - 13",
"Oct\r\n \r\n 8 - 11", "Oct 29 - Nov 1",
"Nov\r\n \r\n 19 - 22", "Jan\r\n \r\n 21 - 24",
"Mar\r\n \r\n 4 - 7", "Apr 29 - May 2"),
points = c("500", "500", "500", "500", "500", "550", "500"
)), row.names = c(1L, 5L, 8L, 11L, 16L, 23L, 32L), class = "data.frame")