Question

我的数据框中有2列，我想要执行添加小时或添加天数等操作。

例如，我想检查我的date2是否在[date1; date1 + 30天]

我试过这个：

  table(data$date1 <= data$date2 & data$date2 <= data$date1 + 30)

我收到了这个错误：

  Error in data$date1 + 30 : 
  non-numeric argument to binary operator

我将日期格式的date1和date2转换为：

  data$date1 <- substr(data$date1,1,16)
  data$date1 <- format(data$date1, format="%Y-%m-%d %H:%M")

以下是我的数据主管：

     id1   id2            date2              date1
  1 CD0H   15741        2012/02/08 10:03    16/02/2015 16:22
  2 CD00   15058        2011/05/19 09:25    07/05/2015 10:39
  3 CHY0   15987        2011/01/20 11:58    06/02/2015 14:11
  4 CTPO   15254        2010/09/29 12:45    01/04/2015 04:49
  5 CDHY   15051                            06/05/2015 15:01
  6 CDJU   15035                            17/04/2015 08:56

Answer 1

或许转换为日期/时间对象不起作用：

def findAinD(cookies, cream):  # assumes that cookies and cream can be treated as such in the for loop will fail otherwise
    A1 = []
    D1 = []

    for mrow in range(1, Cookies.get_highest_row() + 1):
        A1 += cookies['A' + str(mrow)]
        D1 += cream['D' + str(mrow)]

    A1.sort()  # Alphabetical
    D1.sort()  # ^


    for i, cookie in enumerate(A1): # Enumerate returns the index and the object for each iteration
        A1[i] = D1.index(cookie)    # If cookie IS in D, then A1[i] now contains the index of the first occurence of A[i] in D
                                    # If cookie is not, then the result is -1, which is never an index,
                                    #  and we filter those out before round 2 (not shown)

    return A1

班级仍然是“角色”：

library( lubridate )

data <- read.table( filename, header=TRUE, sep = ";" )

data$date1 <- substr(data$date1,1,16)
data$date1 <- format(data$date1, format="%Y/%m/%d %H:%M")

data$date2 <- substr(data$date2,1,16)
data$date2 <- format(data$date2, format="%d/%m/%Y %H:%M")

我按如下方式进行了转换：

> class(data$date1)
[1] "character"

这是更复杂的，但至少我们有时间和日期：

library( timeDate )

table <- read.table( filename, header=TRUE, sep = ";",
                     colClasses = c( "factor", "numeric", "character", "character" ))

data <- cbind( table[1:2],
               apply( table[3], 2, FUN=function(x){ timeDate(x,format="%Y/%m/%d %H:%M") } ),
               apply( table[4], 2, FUN=function(x){ timeDate(x,format="%d/%m/%Y %H:%M") } ) )

colnames(data) <- colnames(table)

Answer 2

查看lubridate库以了解日期和时间。

以下是使用lubridate添加30天的方法

library(lubridate)

ymd('2011-01-01') + days(30)
[1] "2011-01-31 UTC"

了解更多信息：

http://cran.r-project.org/web/packages/lubridate/vignettes/lubridate.html

如何在日期执行操作，如“％Y /％m /％d％H：％M”？

2 个答案: