113

I have a dataset that looks like this:

Month    count
2009-01  12
2009-02  310
2009-03  2379
2009-04  234
2009-05  14
2009-08  1
2009-09  34
2009-10  2386

I want to plot the data (months as x values and counts as y values). Since there are gaps in the data, I want to convert the Information for the Month into a date. I tried:

as.Date("2009-03", "%Y-%m")

But it did not work. Whats wrong? It seems that as.Date() requires also a day and is not able to set a standard value for the day? Which function solves my problem?

AndrewGB
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R_User
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9 Answers9

97

Since dates correspond to a numeric value and a starting date, you indeed need the day. If you really need your data to be in Date format, you can just fix the day to the first of each month manually by pasting it to the date:

month <- "2009-03"
as.Date(paste(month, "-01", sep=""))
ah bon
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Sacha Epskamp
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  • What other formats for dates are there? I saw something with POSIX and something with ISO, but I'm not sure if those are different formats. I thought those are just functions,... – R_User Jun 05 '11 at 12:57
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    Worth noting that you can specify the day as being the same in the formatter, so you can do `as.Date(month, format='%Y-%m-01')` and achieve the same outcome. This "feels" preferable to me since specifying the same date in each month is more about the format of the date then string manipulation, but maybe that is nonsense. – JBecker May 24 '13 at 16:57
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    @JBecker your suggestion doesn't work for me. `> as.Date("2016-01", format="%Y-%m-01") # [1] NA`. I'm using R 3.3.1 – n8sty Jan 20 '17 at 20:43
65

Try this. (Here we use text=Lines to keep the example self contained but in reality we would replace it with the file name.)

Lines <- "2009-01  12
2009-02  310
2009-03  2379
2009-04  234
2009-05  14
2009-08  1
2009-09  34
2009-10  2386"

library(zoo)
z <- read.zoo(text = Lines, FUN = as.yearmon)
plot(z)

The X axis is not so pretty with this data but if you have more data in reality it might be ok or you can use the code for a fancy X axis shown in the examples section of ?plot.zoo .

The zoo series, z, that is created above has a "yearmon" time index and looks like this:

> z
Jan 2009 Feb 2009 Mar 2009 Apr 2009 May 2009 Aug 2009 Sep 2009 Oct 2009 
      12      310     2379      234       14        1       34     2386

"yearmon" can be used alone as well:

> as.yearmon("2000-03")
[1] "Mar 2000"

Note:

  1. "yearmon" class objects sort in calendar order.

  2. This will plot the monthly points at equally spaced intervals which is likely what is wanted; however, if it were desired to plot the points at unequally spaced intervals spaced in proportion to the number of days in each month then convert the index of z to "Date" class: time(z) <- as.Date(time(z)) .

G. Grothendieck
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37

The most concise solution if you need the dates to be in Date format:

library(zoo)
month <- "2000-03"
as.Date(as.yearmon(month))
[1] "2000-03-01"

as.Date will fix the first day of each month to a yearmon object for you.

Ben Rollert
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32

You could also achieve this with the parse_date_time or fast_strptime functions from the lubridate-package:

> parse_date_time(dates1, "ym")
[1] "2009-01-01 UTC" "2009-02-01 UTC" "2009-03-01 UTC"

> fast_strptime(dates1, "%Y-%m")
[1] "2009-01-01 UTC" "2009-02-01 UTC" "2009-03-01 UTC"

The difference between those two is that parse_date_time allows for lubridate-style format specification, while fast_strptime requires the same format specification as strptime.

For specifying the timezone, you can use the tz-parameter:

> parse_date_time(dates1, "ym", tz = "CET")
[1] "2009-01-01 CET" "2009-02-01 CET" "2009-03-01 CET"

When you have irregularities in your date-time data, you can use the truncated-parameter to specify how many irregularities are allowed:

> parse_date_time(dates2, "ymdHMS", truncated = 3)
[1] "2012-06-01 12:23:00 UTC" "2012-06-01 12:00:00 UTC" "2012-06-01 00:00:00 UTC"

Used data:

dates1 <- c("2009-01","2009-02","2009-03")
dates2 <- c("2012-06-01 12:23","2012-06-01 12",'2012-06-01")
Jaap
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  • having converted a character variable to format `date` using `parse_date_time`, is there a way to view it in a different order than `"2009-01-01 UTC"` using `lubridate` package? I would prefer to see the day first in my dataset e.g. `01-01-2009`. – user63230 Jul 30 '19 at 15:34
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    @user63230 See `?format`; e.g.: `format(your_date, "%d-%m-%Y")`. There is a disadvantage to this though: you wil get a character value back and not a date. – Jaap Jul 30 '19 at 20:18
  • Thanks but I was trying to avoid `format` for the reason you mention, I thought there might be a way to incorporate this in the `lubridate` package but seems there isn't. – user63230 Jul 31 '19 at 08:38
15

Using anytime package:

library(anytime)

anydate("2009-01")
# [1] "2009-01-01"
zx8754
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  • That's a little weird that it chooses "01-01", is there anything in the documentation about the choice? Maybe more illustrative to also show `anydate("2009-03")` if it always chooses the first day of the month. – lmo Sep 01 '17 at 18:01
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    @lmo didn't check the docs, I'd say this is "common" practice when dd is missing to choose 1st day. – zx8754 Sep 01 '17 at 18:47
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    That makes sense. I was vaguely remembered and then found what triggered the comment. From the Note section of `?strptime`: *the input string need not specify the date completely: it is assumed that unspecified seconds, minutes or hours are zero, and an unspecified year, month or day is the current one. (However, if a month is specified, the day of that month has to be specified by %d or %e since the current day of the month need not be valid for the specified month.)* It looks like megatron's answer contains a similar piece of documentation from `as.Date`. – lmo Sep 01 '17 at 18:52
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    for years before 1900, it does not work. For example, I tried this `anytime('1870-01')` – msh855 Apr 13 '19 at 13:04
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    Only solution that worked for me! use anydate() instead of anytime() – schroederadrian Nov 07 '21 at 16:50
6

Indeed, as has been mentioned above (and elsewhere on SO), in order to convert the string to a date, you need a specific date of the month. From the as.Date() manual page:

If the date string does not specify the date completely, the returned answer may be system-specific. The most common behaviour is to assume that a missing year, month or day is the current one. If it specifies a date incorrectly, reliable implementations will give an error and the date is reported as NA. Unfortunately some common implementations (such as glibc) are unreliable and guess at the intended meaning.

A simple solution would be to paste the date "01" to each date and use strptime() to indicate it as the first day of that month.

For those seeking a little more background on processing dates and times in R:

In R, times use POSIXct and POSIXlt classes and dates use the Date class.

Dates are stored as the number of days since January 1st, 1970 and times are stored as the number of seconds since January 1st, 1970.

So, for example:

d <- as.Date("1971-01-01")
unclass(d)  # one year after 1970-01-01
# [1] 365

pct <- Sys.time()  # in POSIXct
unclass(pct)  # number of seconds since 1970-01-01
# [1] 1450276559
plt <- as.POSIXlt(pct)
up <- unclass(plt)  # up is now a list containing the components of time
names(up)
# [1] "sec"    "min"    "hour"   "mday"   "mon"    "year"   "wday"   "yday"   "isdst"  "zone"  
# [11] "gmtoff"
up$hour
# [1] 9

To perform operations on dates and times:

plt - as.POSIXlt(d)
# Time difference of 16420.61 days

And to process dates, you can use strptime() (borrowing these examples from the manual page):

strptime("20/2/06 11:16:16.683", "%d/%m/%y %H:%M:%OS")
# [1] "2006-02-20 11:16:16 EST"

# And in vectorized form:
dates <- c("1jan1960", "2jan1960", "31mar1960", "30jul1960")
strptime(dates, "%d%b%Y")
# [1] "1960-01-01 EST" "1960-01-02 EST" "1960-03-31 EST" "1960-07-30 EDT"
Megatron
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2

I think @ben-rollert's solution is a good solution.

You just have to be careful if you want to use this solution in a function inside a new package.

When developping packages, it's recommended to use the syntaxe packagename::function_name() (see http://kbroman.org/pkg_primer/pages/depends.html).

In this case, you have to use the version of as.Date() defined by the zoo library.

Here is an example : 

> devtools::session_info()
Session info ----------------------------------------------------------------------------------------------------------------------------------------------------
 setting  value                       
 version  R version 3.3.1 (2016-06-21)
 system   x86_64, linux-gnu           
 ui       RStudio (1.0.35)            
 language (EN)                        
 collate  C                           
 tz       <NA>                        
 date     2016-11-09                  

Packages --------------------------------------------------------------------------------------------------------------------------------------------------------

 package  * version date       source        
 devtools   1.12.0  2016-06-24 CRAN (R 3.3.1)
 digest     0.6.10  2016-08-02 CRAN (R 3.2.3)
 memoise    1.0.0   2016-01-29 CRAN (R 3.2.3)
 withr      1.0.2   2016-06-20 CRAN (R 3.2.3)

> as.Date(zoo::as.yearmon("1989-10", "%Y-%m")) 
Error in as.Date.default(zoo::as.yearmon("1989-10", "%Y-%m")) : 
  do not know how to convert 'zoo::as.yearmon("1989-10", "%Y-%m")' to class “Date”

> zoo::as.Date(zoo::as.yearmon("1989-10", "%Y-%m"))
[1] "1989-10-01"

So if you're developping a package, the good practice is to use : 

zoo::as.Date(zoo::as.yearmon("1989-10", "%Y-%m"))
PAC
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1

A way using ym from lubridate.

The month can either be a number, an abbreviated month or a full month name with a variety of separators (even without separator), e.g.

library(lubridate)

ym(c("2012/September", "2012-Aug", "2012.07", 201204))
[1] "2012-09-01" "2012-08-01" "2012-07-01" "2012-04-01"

on the given data:

ym(dat$Month)
[1] "2009-01-01" "2009-02-01" "2009-03-01" "2009-04-01" "2009-05-01"
[6] "2009-08-01" "2009-09-01" "2009-10-01"

Note that there's also my if you have it the other way round, e.g. Sep/2022.

Data

dat <- structure(list(Month = c("2009-01", "2009-02", "2009-03", "2009-04",
"2009-05", "2009-08", "2009-09", "2009-10"), count = c(12L, 310L,
2379L, 234L, 14L, 1L, 34L, 2386L)), class = "data.frame", row.names = c(NA,
-8L))
Andre Wildberg
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0

tidyverse recently added the clock package in addition to lubridate that has some nice functionality for this:

library(clock)

x <- year_month_day_parse(df$Month, format = "%Y-%m", precision = "month") 
# <year_month_day<month>[8]>
# [1] "2009-01" "2009-02" "2009-03" "2009-04" "2009-05" "2009-08" "2009-09" "2009-10"

Date Manipulation and Extraction

The output of this is a year-month-day vector where you can still do date arithmetic and apply other common functions as expected:

sort(x, decreasing = T)
# <year_month_day<month>[8]>
# [1] "2009-10" "2009-09" "2009-08" "2009-05" "2009-04" "2009-03" "2009-02" "2009-01"

add_months(x, 3)
# <year_month_day<month>[8]>
# [1] "2009-04" "2009-05" "2009-06" "2009-07" "2009-08" "2009-11" "2009-12" "2010-01"

add_years(x, -2)
# <year_month_day<month>[8]>
# [1] "2007-01" "2007-02" "2007-03" "2007-04" "2007-05" "2007-08" "2007-09" "2007-10"

get_month(x)
# [1]  1  2  3  4  5  8  9 10

You can also set the day, if you need it, with set_day:

set_day(x, 1)
<year_month_day<day>[8]>
[1] "2009-01-01" "2009-02-01" "2009-03-01" "2009-04-01" "2009-05-01" "2009-08-01"
[7] "2009-09-01" "2009-10-01"

Handling Invalid Dates

Or if you wanted to cleanly get the last day of every month with this structure, the invalid_* set of functions can help:

# not 31 days in Feb, Apr, Sep
y <- set_day(x, 31)
# <year_month_day<day>[8]>
# [1] "2009-01-31" "2009-02-31" "2009-03-31" "2009-04-31" "2009-05-31" "2009-08-31"
# [7] "2009-09-31" "2009-10-31"

invalid_any(y)
[1] TRUE

invalid_detect(y)
[1] FALSE  TRUE FALSE  TRUE FALSE FALSE  TRUE FALSE

You can handle invalid dates with invalid_resolve or you can use drop them with invalid_remove:

invalid_resolve(y, invalid = "previous")
<year_month_day<day>[8]>
[1] "2009-01-31" "2009-02-28" "2009-03-31" "2009-04-30" "2009-05-31" "2009-08-31"
[7] "2009-09-30" "2009-10-31"

From the documentation you can specify the following values for the invalid argument to handle invalid dates:

"previous": The previous valid instant in time.

"previous-day": The previous valid day in time, keeping the time of day.

"next": The next valid instant in time.

"next-day": The next valid day in time, keeping the time of day.

"overflow": Overflow by the number of days that the input is invalid by. Time of day is dropped.

"overflow-day": Overflow by the number of days that the input is invalid by. Time of day is kept.

"NA": Replace invalid dates with NA.

"error": Error on invalid dates.

LMc
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