Question

我有以下数据框：

Date_from <- c("2013-01-01","2013-01-04")
Date_to <- c("2013-01-03","2013-01-06")
Parameter <- c("Par1","Par1","Par2","Par2")
conc<-c("1.5","2.5","1.5","1.8")
metals<-data.frame(Date_from,Date_to,Parameter,conc)
metals$Date_from<-as.Date(metals$Date_from)
metals$Date_to<-as.Date(metals$Date_to)
metals$conc<-as.numeric(as.character(metals$conc))

我需要做的是为每个参数提取每个日期范围内的日期，并将浓度值分配给该范围内的每个日期，并将所有这些信息放入新的数据框中。结果应如下所示：

Date        Parameter    conc
2013-01-01  Par1         1.5
2013-01-02  Par1         1.5
2013-01-03  Par1         1.5
2013-01-04  Par1         2.5
2013-01-05  Par1         2.5
2013-01-06  Par1         2.5
2013-01-01  Par2         1.5
2013-01-02  Par2         1.5
2013-01-03  Par2         1.5
2013-01-04  Par2         1.8
2013-01-05  Par2         1.8
2013-01-06  Par2         1.8

Answer 1

这是list的一个选项。通过将{Date_from'到{Date_to'（seq）的map来创建select列，删除不需要的列（unnest）和library(tidyverse) metals %>% mutate(Date = map2(Date_from, Date_to, seq, by = "1 day")) %>% select(-Date_from, -Date_to) %>% unnest %>% select(Date, Parameter, conc) # Date Parameter conc #1 2013-01-01 Par1 1.5 #2 2013-01-02 Par1 1.5 #3 2013-01-03 Par1 1.5 #4 2013-01-04 Par1 2.5 #5 2013-01-05 Par1 2.5 #6 2013-01-06 Par1 2.5 #7 2013-01-01 Par2 1.5 #8 2013-01-02 Par2 1.5 #9 2013-01-03 Par2 1.5 #10 2013-01-04 Par2 1.8 #11 2013-01-05 Par2 1.8 #12 2013-01-06 Par2 1.8

base R

或者可以通过lst <- Map(seq, MoreArgs = list(by = "1 day"), metals$Date_from, metals$Date_to) out <- cbind(Date = do.call(c, lst), metals[rep(seq_len(nrow(metals)), lengths(lst)), c("Parameter", "conc")]) row.names(out) <- NULL out # Date Parameter conc #1 2013-01-01 Par1 1.5 #2 2013-01-02 Par1 1.5 #3 2013-01-03 Par1 1.5 #4 2013-01-04 Par1 2.5 #5 2013-01-05 Par1 2.5 #6 2013-01-06 Par1 2.5 #7 2013-01-01 Par2 1.5 #8 2013-01-02 Par2 1.5 #9 2013-01-03 Par2 1.5 #10 2013-01-04 Par2 1.8 #11 2013-01-05 Par2 1.8 #12 2013-01-06 Par2 1.8

完成

try {} catch() {}

Answer 2

我们可以在不依赖57个软件包的情况下做到这一点：

metals <- data.frame(Date_from,Date_to,Parameter,conc)

do.call(
  rbind.data.frame,
  lapply(1:nrow(metals), function(.i) {
    data.frame(
      Date = seq(as.Date(metals$Date_from[.i]), as.Date(metals$Date_to[.i]), "1 day"),
      Parameter = metals$Parameter[.i],
      conc = as.double(as.character(metals$conc[.i])),
      stringsAsFactors = FALSE
    )
  })
)

使用来自OP的经过预类型转换的数据帧：

library(microbenchmark)

microbenchmark(
  base = do.call(
    rbind.data.frame,
    lapply(1:nrow(metals), function(.i) {
      data.frame(
        Date = seq(metals$Date_from[.i], metals$Date_to[.i], "1 day"),
        Parameter = metals$Parameter[.i],
        conc = metals$conc[.i],
        stringsAsFactors = FALSE
      )
    })
  ),
  base2 = {
    lst <- Map(
      seq, MoreArgs = list(by = "1 day"), metals$Date_from, metals$Date_to
    )
    cbind(
      Date = do.call(c, lst), 
      metals[rep(seq_len(nrow(metals)), lengths(lst)), c("Parameter", "conc")]
    )
  },
  tidy = metals %>% 
    mutate(Date = map2(Date_from, Date_to, seq, by = "1 day")) %>% 
    select(-Date_from, -Date_to) %>%
    unnest %>%
    select(Date, Parameter, conc)
)
## Unit: microseconds
##   expr      min        lq      mean    median        uq       max neval
##   base 2472.997 2615.7025 2758.6086 2678.6220 2765.6375  8085.012   100
##  base2  716.680  784.0505  835.0233  815.9715  869.8095  1166.096   100
##   tidy 7331.729 7671.4065 8644.6002 7889.7080 8080.5925 82376.963   100

从日期范围中提取日期并分配值

2 个答案: