Problems Formatting Date into Format "%Y-%M"

Problems formatting date into format %Y-%m

A year and a month do not make a date. You need a day also.

d <- data.frame(V1=c("1950-12","1951-01"))
as.Date(paste(d$V1,1,sep="-"),"%Y-%m-%d")
# [1] "1950-12-01" "1951-01-01"

You could also use the yearmon class in the zoo package.

library(zoo)
as.yearmon(d$V1)
# [1] "Dec 1950" "Jan 1951"

Convert dates to %y-%m-%d format in Python

Step 0:-

Your dataframe:-

df=pd.read_csv('your file name.csv')

Step 1:-

firstly convert your 'date' column into datetime by using to_datetime() method:-

df['date']=pd.to_datetime(df['date'])

Step 2:-

And If you want to convert them in string like format Then use:-

df['date']=df['date'].astype(str)

Now if you print df or write df(if you are using jupyter notebook)

Output:-

0    2020-01-01
1    2020-12-31
2    2020-06-20

Format JavaScript date as yyyy-mm-dd

You can do:

function formatDate(date) {
    var d = new Date(date),
        month = '' + (d.getMonth() + 1),
        day = '' + d.getDate(),
        year = d.getFullYear();

    if (month.length < 2) 
        month = '0' + month;
    if (day.length < 2) 
        day = '0' + day;

    return [year, month, day].join('-');
}
 
console.log(formatDate('Sun May 11,2014'));

How to format a UTC date as a `YYYY-MM-DD hh:mm:ss` string using NodeJS?

If you're using Node.js, you're sure to have EcmaScript 5, and so Date has a toISOString method. You're asking for a slight modification of ISO8601:

new Date().toISOString()
> '2012-11-04T14:51:06.157Z'

So just cut a few things out, and you're set:

new Date().toISOString().
  replace(/T/, ' ').      // replace T with a space
  replace(/\..+/, '')     // delete the dot and everything after
> '2012-11-04 14:55:45'

Or, in one line: new Date().toISOString().replace(/T/, ' ').replace(/\..+/, '')

ISO8601 is necessarily UTC (also indicated by the trailing Z on the first result), so you get UTC by default (always a good thing).

Converting year and month ( yyyy-mm format) to a date?

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)) .

as.Date with dates in format m/d/y in R

Use capital Y in as.Date call instead. This should do the trick:

> as.Date("3/15/2012", "%m/%d/%Y")
[1] "2012-03-15"

From the help file's examples you can realize when year is full specified you should use %Y otherwise %y for example:

> dates <- c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92")
> as.Date(dates, "%m/%d/%y")
[1] "1992-02-27" "1992-02-27" "1992-01-14" "1992-02-28" "1992-02-01"

You can see that in your example the Year format is 2012 then you should use %Y, and in the other example (taken from the as.Date help file) Year format is 92 then using %y is the correct way to go. See as.Date for further details.

How to convert Python date format '%B - %Y' back to '%Y-%m-%d'?

Edit ::

I didn't do any benchmark, but just so you know, casting your column as date, doing... df['your_column'] = pd.to_datetime(df['your_column']) will convert any date you have to an ISO format. See below the second example.

The second example should be way faster though. :)

import pandas as pd
from datetime import datetime

data = {
    'A' : ['July - 2019', 'June - 2020'],
    'B' : [1, 2]
}

df = pd.DataFrame(data)
print(df, end='\n\n')
#              A  B
# 0  July - 2019  1
# 1  June - 2020  2

day_to_put = 15
df['A'] = df['A'].apply( lambda x: datetime.strptime(x, '%B - %Y')\
                                           .replace(day=day_to_put)\
                                           .strftime('%Y-%m-%d') )

print(df)
#            A  B
#0  2019-07-15  1
#1  2020-06-15  2

Second example

import pandas as pd
from datetime import datetime

data = {
    'A' : ['July - 2019', 'June - 2020'],
    'B' : [1, 2]
}

df = pd.DataFrame(data)
print(df, end='\n\n')
#              A  B
# 0  July - 2019  1
# 1  June - 2020  2

df['A'] = pd.to_datetime(df['A'])

print(df)
#             A  B
# 0  2019-07-01  1
# 1  2020-06-01  2

Pandas - Datetime format change to '%m/%d/%Y'

The reason you have to use errors="ignore" is because not all the dates you are parsing are in the correct format. If you use errors="coerce" like @phi has mentioned then any dates that cannot be converted will be set to NaT. The columns datatype will still be converted to datatime64 and you can then format as you like and deal with the NaT as you want.

Example

A dataframe with one item in Date not written as Year/Month/Day (25th Month is wrong):

>>> df = pd.DataFrame({'ID': [91060, 91061, 91062, 91063], 'Date': ['2017/11/10', '2022/05/01', '2022/04/01', '2055/25/25']})
>>> df
      ID        Date
0  91060  2017/11/10
1  91061  2022/05/01
2  91062  2022/04/01
3  91063  2055/25/25

>>> df.dtypes
ID       int64
Date    object
dtype: object

Using errors="ignore":

>>> df['Date'] = pd.to_datetime(df['Date'], errors='ignore')
>>> df
      ID        Date
0  91060  2017/11/10
1  91061  2022/05/01
2  91062  2022/04/01
3  91063  2055/25/25

>>> df.dtypes
ID       int64
Date    object
dtype: object

Column Date is still an object because not all the values have been converted. Running df['Date'] = df['Date'].dt.strftime("%m/%d/%Y") will result in the AttributeError

Using errors="coerce":

>>> df['Date'] = pd.to_datetime(df['Date'], errors='coerce')
>>> df
      ID       Date
0  91060 2017-11-10
1  91061 2022-05-01
2  91062 2022-04-01
3  91063        NaT

>>> df.dtypes
ID               int64
Date    datetime64[ns]
dtype: object

Invalid dates are set to NaT and the column is now of type datatime64 and you can now format it:

>>> df['Date'] = df['Date'].dt.strftime("%m/%d/%Y")
>>> df
      ID        Date
0  91060  11/10/2017
1  91061  05/01/2022
2  91062  04/01/2022
3  91063         NaN

Note: When formatting datatime64, it is converted back to type object so NaT's are changed to NaN. The issue you are having is a case of some dirty data not in the correct format.

Dealing with +00:00 in datetime format

The easiest thing to do is let pd.to_datetime auto-infer the format. That works very well for standard formats like this (ISO 8601):

import pandas as pd

dti = pd.to_datetime(["2020-06-30 15:20:13.078196+00:00"])

print(dti)
# DatetimeIndex(['2020-06-30 15:20:13.078196+00:00'], dtype='datetime64[ns, UTC]', freq=None)

+00:00 is a UTC offset of zero hours, thus can be interpreted as UTC.

btw., pd.to_datetime also works very well for mixed formats, see e.g. here.

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