I'm converting an string type of Time series to datetime in Python and I'm so confused that why is my datetime always display the result I don't expect. \n
what I want is shown in my img here
import datetime
time = '23:30:00' # Time in string format
dt=datetime.datetime.strptime(time, '%H:%M:%S')
print(dt.time()) # time method will only return the time
I hope this helps
You should put your question in the question, not some off-site illustration. We do have code blocks available. Also, you converted to Pandas datetime, not Python datetime. Both of these have "date" in their name because they do contain the date. You could represent just a time using e.g. Pandas timedelta or Python datetime.time. The format you pass to panads.to_datetime is how to parse the input, not how to display the result.
You have converted your string Series to a Series of pd.Timestamp. Internally a Timestamp is a number of nanoseconds from 1970-01-01 00:00:00.
The correct way to format a date in pandas is to convert it to a string with .dt.strfime, *when you no longer need to process it as a Timestamp.
TL/DR:
if you want it in HH:MM:SS format leave it in string dtype
if you need to process it as a Timestampand yet have it in HH:MM:SS format, convert it to Timestamp, process it and when done convert it back to a string
Related
This is my data :
dates = np.arange("2018-01-01", "2021-12-31", dtype="datetime64[D]")
I now want to convert from :
"2018-01-01" -> "Jan-01-2018" ["Monthname-day-year"] format
How to i do this ?
Is it possible to initialize this in the way we want to convert ?
Can i use something like:
for i in dates:
i = i.replace(i.month,i.strftime("%b"))
You can try this:
from datetime import datetime
dates = np.arange("2018-01-01", "2021-12-31", dtype="datetime64[D]")
result_dates = []
for date in dates.astype(datetime):
result_dates.append(date.strftime("%b-%d-%Y"))
But you will need to convert result dates as shown in the code
I feel compelled to elaborate on Silvio Mayolo's very relevant but ostensibly ignored comment above. Python stores a timestamp as structure (see How does Python store datetime internally? for more information) Hence, the DateTime does not as such have a 'format'. A format only becomes necessary when you want to print the date because you must first convert the timestamp to a string. Thus, you do NOT need to initialise any format. You only need to declare a format when the time comes to print the timestamp.
While you CAN store the date as a string in your dataframe index in a specific format, you CANNOT perform time related functions on it without first converting the string back to a time variable. ie current_time.hour will return an integer with the current hour if current_time is a datetime variable but will crash if it is a string formatted as a timestamp (such as "2023-01-15 17:23").
This is important to understand, because eventually you will need to manipulate the variables and need to understand whether you are working with a time or a string.
I have a column of float values which are tweet creation dates. This is the code I used to convert them from float to datetime:
t = 1508054212.0
datetime.utcfromtimestamp(t).strftime('%Y-%m-%d %H:%M:%S')
All the values returned belong to October 2017. However, the data is supposed to be collected over multiple months. So the dates should have different months and not just different Hours, Minutes and Seconds.
These are some values which I need to convert:
1508054212.0
1508038548.0
1506890436.0
Request you to suggest an alternative approach to determine the dates. Thank you.
I assumed df['tweet_creation'].loc[1] will return a number like the examples you gave.
Unfortunately, I don't know what f is, but I assumed it was a float.
My answer is inspired by this other answer: Converting unix timestamp string to readable date. You have a UNIX timestamp, so the easiest way is to use it and not convert it as a string.
from datetime import datetime, timedelta
dtobj = datetime.utcfromtimestamp(int(df['tweet_creation'].loc[1])) + timedelta(days=f-int(f))
To have the string representation you can use the function strftime.
I'm working with some video game speedrunning (basically, races where people try to beat a game as fast as they can) data, and I have many different run timings in HH:MM:SS format. I know it's possible to convert to seconds, but I want to keep in this format for the purposes of making the axes on any graphs easy to read.
I have all the data in a data frame already and tried converting the timing data to datetime format, with format = '%H:%M:%S', but it just uses this as the time on 1900-01-01.
data=[['Aggy','01:02:32'], ['Kirby','01:04:54'],['Sally','01:06:04']]
df=pd.DataFrame(data, columns=['Runner','Time'])
df['Time']=pd.to_datetime(df['Time'], format='%H:%M:%S')
I thought specifying the format to be just hours/minutes/seconds would strip away any date, but when I print out the header of my dataframe, it says that the time data is now 1900-01-01 01:02:32, as an example. 1:02:32 AM on January 1st, 1900. I want Python to recognize the 1:02:32 as a duration of time, not a datetime format. What's the best way to go about this?
The format argument defines the format of the input date, not the format of the resulting datetime object (reference).
For your needs you can either use the H:m:s part of the datetime, or use the to_timedelta
method.
I am trying to use timestamp information to graph.
However, when I convert the number into hh:mm:ss. It does not work.
I have tried this:
timestamp = [dt.strptime(str(datetime.timedelta(seconds=round(t/1000))),'%H:%M:%S') for t in timestamp1]
Also I tried this
timestamp = [dt.strftime(dt.strptime(str(datetime.timedelta(seconds=round(t / 1000))), '%H:%M:%S'), '%H:%M:%S') for t in timestamp]
However, it is possible to see the list with the new values. However, I have problems with the graphs and these new values.
Can anybody help me?
Try using this and make sure to import the time module!
timestamp = [time.strftime('%H:%M:%S', time.gmtime(t/1000)) for t in timestamp1]
In your code you're using datetime.strptime(date_string, format) function, it's main use is to parse time strings in order to create datetime object.
In order to get a parsed string of a desired format you should use date.strftime(format) function instead.
So basically in your code you can only add it and should get the wanted result:
import datetime
from datetime import datetime as dt
timestamp = [dt.strftime(dt.strptime(str(datetime.timedelta(seconds=round(t / 1000))), '%H:%M:%S'), '%H:%M:%S') for t in timestamp1]
I'm not sure what was your input and desired output, but you could also use date.fromtimestamp(timestamp) for the parsing of timestamps
How can I convert the following time format:
hhmmss.ff (like 110241.22 is 11:02:41.22)
into the date/time format with pandas?
I tries to use pandas.to_datetime() but it fails to do the conversion. Here is an example:
hhmmss='110241.22'
pd.to_datetime(hhmmss)
Thanks
You need to specify the format you want to convert the time to. Here's a helpful resource for figuring out what each symbol means. Here's Pandas documentation
pd.to_datetime(df['column_name'], format = '%H%M%S.%f')