Convert R Character to Date

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.

How to convert character to date with two different types of date formats in R?

You can do something like:

format_ymd  <- as.Date(dates, format = "%Y-%m-%d")
format_dmy  <- as.Date(dates, format = "%d/%m/%Y")
as.Date(ifelse(is.na(format_ymd), format_dmy, format_ymd), origin = "1970-01-01")
# [1] "2022-04-08" "2021-01-26" "2021-07-14" "2021-12-27"

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"

Convert Character to date

how about just overwriting that column as a date column:

library(tidyverse)
data <- data %>% mutate(InvoiceDate = dmy_hm(InvoiceDate)) #dmy_hm() converts to date
#or if you really need a separate vector:
data1 <- dmy_hm(data$InvoiceDate)

you can read about dmy_hm()and similar functions in the tidyverse's sublibrary lubridate. I find it very useful.

How to convert to character to date in R?

If you have locale same as the data that you have using lubridate's mdy_hms should work.

df$date <- lubridate::mdy_hms(df$date)

How to convert character to Date format in R?

Since you have only year and month, you need to assign some value for day before you convert to date. In the example below, the day has arbitrarily been chosen as 15.

IF THE INPUT IS CHARACTER

dates = c("2012.10", "2012.01")

lubridate::ymd(paste0(year_month = dates, day = "15"))
#[1] "2012-10-15" "2012-01-15"

IF THE INPUT IS NUMERIC

dates = c(2012.10, 2012.01)

do.call(c, lapply(strsplit(as.character(dates), "\\."), function(d){
    if(nchar(d[2]) == 1){
        d[2] = paste0(d[2],"0")
    }
    lubridate::ymd(paste0(year = d[1], month = d[2], day = "15"))
}))
#[1] "2012-10-15" "2012-01-15"

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))

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