Converting a Character String into a Date in R

Convert date-time string to class Date

You may be overcomplicating things, is there any reason you need the stringr package? You can use as.Date and its format argument to specify the input format of your string.

 df <- data.frame(Date = c("10/9/2009 0:00:00", "10/15/2009 0:00:00"))
 as.Date(df$Date, format =  "%m/%d/%Y %H:%M:%S")
 # [1] "2009-10-09" "2009-10-15"

Note the Details section of ?as.Date:

Character strings are processed as far as necessary for the format specified: any trailing characters are ignored

Thus, this also works:

as.Date(df$Date, format =  "%m/%d/%Y")
# [1] "2009-10-09" "2009-10-15"

All the conversion specifications that can be used to specify the input format are found in the Details section in ?strptime. Make sure that the order of the conversion specification as well as any separators correspond exactly with the format of your input string.

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More generally and if you need the time component as well, use as.POSIXct or strptime:

as.POSIXct(df$Date, "%m/%d/%Y %H:%M:%S")    
strptime(df$Date, "%m/%d/%Y %H:%M:%S")

I'm guessing at what your actual data might look at from the partial results you give.

Convert string into date format in R

The as.Date() function will convert a string into date format, and the format of the output will always be in yyyy-mm-dd format in R (ISO 8601). The format argument in the as.Date() function is to specify the date format of the string input. I remember I initially thought it was specifying the output format, but it's the input format (you can change the output format with a subsequent format() function, however this will convert it back to a string).

Your string looks to be in ddmmyyyy (%d%m%Y) format, this should be what you specify as the format argument in as.Date(). Your format does not include hyphens, so the format argument should also not include hyphens. Note that ddmmyyyy, dd-mm-yyyy, dd/mm/yyyy, dd.mm.yyyy are all different date formats, even though the day, month and year are in the same order, they would be converted to date with formats %d%m%Y, %d-%m-%Y, %d/%m/%Y, and %d.%m.%Y, respectively.

Further advice on working with dates and times is available in the relevant chapters of R for Data Science by Wickham & Grolemund and The R Cookbook by Teetor & Long.

Convert Character into Date (R)

You could try adding a default day to the date, e.g. 01, to make it a formal date:

data$year <- substr(data$month, 1, 4)
data$mon <- substr(data$month, 6, 8)
data$date <- as.Date(paste(data$year, data$mon, "01", sep="-"))

How to convert character into date format in R

A base R option (assuming that there are only two formats in the OP's 'create_date' column), will be to create a logical index with grepl for those date elements that start with 'year', subset the 'create_date' based on the logical index ('i1'), convert to Date class separately and assign those separately to a Date vector of the same length as the number of rows of the dataset to create the full Date class.

i1 <- grepl("^[0-9]{4}", df$create_date)
v1 <- as.Date(df$create_date[i1])
v2 <- as.Date(df$create_date[!i1],  "%m/%d/%Y")
v3 <- Sys.Date() + 0:(nrow(df)-1)   
v3[i1] <- v1
v3[!i1] <- v2
df$create_date <- v3

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Or as I commented in the OP's post (first) parse_date_time from lubridate can be used

library(lubridate)
as.Date(parse_date_time(df$create_date, c('mdy', 'ymd_hms')))
#[1] "2016-08-13" "2016-08-13" "2016-08-13" "2016-08-13" 
#[5] "2016-08-13" "2016-08-13" "2016-08-13"

data

df <- structure(list(create_date = c("8/13/2016", "8/13/2016", 
"8/13/2016", 
"2016-08-13T08:26:04Z", "2016-08-13T14:30:23Z", "8/13/2016", 
"8/13/2016")), .Names = "create_date", class = "data.frame",
 row.names = c(NA, -7L))

Convert an entire column from character to date class in R

Replacing the '.' with '/' works from your first example in the format:

Date <- mutate(crime, ArrestDate = as.Date(ArrestDate, format= "%m/%d/%Y"))
class(Date$ArrestDate)
[1] "Date"

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