In Python 3, I would like to change 2018-01-01 to "January first, two thousand and eighteen". I looked at the format guide in the Python datetime documentation
but I did not see a format for spelling out the day of the month or for spelling out the year. Is there a well-known module that can do this (I am new to Python)?
Using the inflect module, as commented by # Lakshay Garg you can do:
import datetime as dt, inflect
p = inflect.engine()
def date_in_words(x):
a = dt.datetime.strptime(x,'%Y-%m-%d')
return (a.strftime('%B ')+p.number_to_words(p.ordinal(int(a.day)))+', ' +
p.number_to_words(int(a.year)))
date_in_words('2018-03-14')
Out[259]: 'March fourteenth, two thousand and eighteen'
Related
I have a date variable calls today_date as below. I need to get the 1st calendar day of the current and next month.
In my case, today_date is 4/17/2021, I need to create two more variables calls first_day_current which should be 4/1/2021, and first_day_next which should be 5/1/2021.
Any suggestions are greatly appreciated
import datetime as dt
today_date
'2021-04-17'
Getting just the first date of a month is quite simple - since it equals 1 all the time. You can even do this without needing the datetime module to simplify calculations for you, if today_date is always a string "Year-Month-Day" (or any consistent format - parse it accordingly)
today_date = '2021-04-17'
y, m, d = today_date.split('-')
first_day_current = f"{y}-{m}-01"
y, m = int(y), int(m)
first_day_next = f"{y+(m==12)}-{m%12+1}-01"
If you want to use datetime.date(), then you'll anyway have to convert the string to (year, month, date) ints to give as arguments (or do today_date = datetime.date.today().
Then .replace(day=1) to get first_day_current.
datetime.timedelta can't add months (only upto weeks), so you'll need to use other libraries for this. But it's more imports and calculations to do the same thing in effect.
I found out pd.offsets could accomplish this task as below -
import datetime as dt
import pandas as pd
today_date #'2021-04-17' this is a variable that is being created in the program
first_day_current = today_date.replace(day=1) # this will be 2021-04-01
next_month = first_day_current + pd.offsets.MonthBegin(n=1)
first_day_next = next_month.strftime('%Y-%m-%d') # this will be 2021-05-01
I've got lots of dates that look like this: 16.8.18 (American: 8/16/18) of type string. Now, I need to check if a date is in the past or future but, datetime doesn't support the German format.
How can I accomplish this?
from datetime import datetime
s = "16.8.18"
d = datetime.strptime(s, "%d.%m.%y")
if d > datetime.now():
print('Date is in the future.')
else:
print('Date is in the past.')
Prints (today is 20.7.2018):
Date is in the future.
The format used in strptime() is explained in the manual pages.
I am working on my project on a timetable app. I can get the date with:
import datetime
now = datetime.datetime.now()
print (now.day, "/", now.month, "/", now.year)
But I can't get the day of the week though. Can someone help me?
To format and print dates, you should use the strftime functions (see the strftime python 3 documentation) instead of manually build your own format.
so e.g.
import datetime
now = datetime.datetime.now()
print(now.strftime("%A, %d/%m/%Y"))
Check out the doc, for the full list of styles. Maybe you want %a (abbreviated weekday name, or also %b or %B for the month name.
If you need just the values, check the datetime documenation, in the same page: you have now.weekday() (Monday is 0 and Sunday is 6), or now.iweekday() (Monday is 1 and Sunday is 7).
Try this:
import time
localtime = time.localtime(time.time())
current_time = time.asctime(localtime)
print(current_time[:3])
This should work.
Thanks.
There is an excellent answer by #seddonym on https://stackoverflow.com/a/29519293/4672536 to find the day of the week. I will post the code here for reference. Good luck! :) :
>>> from datetime import date
>>> import calendar
>>> my_date = date.today()
>>> calendar.day_name[my_date.weekday()]
'Wednesday'
Suppose i have a txt. file that looks like this:
0 day0 event_data0
1 day1 event_data1
2 day2 event_data2
3 day3 event_data3
4 day4 event_data4
........
n dayn event_datan
#where:
#n is the event index
#dayn is the day when the event happened. year-month-day format
#event_datan is what happened at the event.
From this file, i need to create a new one with all the events that happened between two specific dates. like after september the 7th 2003 and before christmas 2006.
Could someone help me this problem? Much appreciated!
Looks like the datetime module is what you'll want. Iterate through the file line by line until the timedelta between the current line's date and your beginning threshold date (Sept 7, 2003 in your example) is positive; stop iterating when you breach Christmas 2006. Load the lines into either a pandas dataframe or numpy array.
Lucas, you can try this:
import re
import os
from datetime import datetime as dt
__date_start__ = dt.strptime('2003-09-07', "%Y-%m-%d").date()
__date_end__ = dt.strptime('2006-12-25', "%Y-%m-%d").date()
f = open('file.txt', 'r').read()
os.remove('events.txt')
for i in f:
date = re.search('\d{4}\-\d{2}-\d{2}',i).group(0)
if date != '':
date_converted = dt.strptime(date, '%Y-%m-%d').date()
if (date_converted > __date_start__) and (date_converted < __date_end__):
open('events.txt', 'a').write(i)
You will change __date_start__ and __date_end__ values to your desire interval, then, the code will search in lines a regex that match with the format of date yyyy-mm-dd. So on, it going to compare in range (date start & end) and, if true, append a events.txt file the content of line.
I assume your file is tab delimited so you can use the pandas package to read it. Just add a the first row with the column names (index, date, event) in your .txt file separated by tab and then read in the data.
df = pandas.read_csv('txt_file.txt', sep='\t', index_col=0)
#index_col=0 just sets your first column as index
After you've done so, follow the steps from this link. That will essentially answer your question on how to select events between two dates by simply using this package. That way you can return a new data frame only with those events you need.
You have not described that you want especially for "after September the 7th 2003 and before Christmas 2006." or you have other options for these two dates ?
if specially for "after september the 7th 2003 and before christmas 2006." then you can get result with regex module in my opinion :
import re
c=r"([0-9]{1,2}\s+)(2003-09-07).+(2006-12-25)\s+\w+"
with open("event.txt","r") as f:
file_data=f.readlines()
regex_search=re.search(c,str(file_data))
print(regex_search.group())
You can also use conditions with group() , or you can use findall() method.
this is my first post on stackoveflow and I'm pretty new to programming especially python. I'm in engineering and am learning python to compliment that going forward, mostly at math and graphing applications.
Basically my question is how do I download csv excel data off a source (in my case stock data from google), and plot only certain rows against the date. For myself I want the date against the close value.
Right now the error message I'm getting is timedata '5-Jul-17' does not match '%d-%m-%Y'
previously I was also getting tuple data does not match
The description of the opened csv data in excel is
[7 columns (Date,Open,High,Low,Close,AdjClose,Volume, and the date is organized as 2017-05-30][1]
I'm sure there are other errors as well unfortunately
I would really be grateful for any help on this,
thank you in advance!
--edit--
Upon fiddling some more I don't think names and dtypes are necessary, when I check the matrix dimensions without those identifiers I get (250L, 6L) which seems right. Now my main problem is coverting the dates to something usable, My error now is strptime only accepts strings, so I'm not sure what to use. (see updated code below)
import matplotlib.pyplot as plt
importnumpy as np
from datetime import datetime
def graph_data(stock):
%getting the data off google finance
data = np.genfromtxt('urlgoeshere'+stock+'forthecsvdata', delimiter=',',
skip_header=1)
# checking format of matrix
print data.shape (returns 250L,6L)
time_format = '%d-%m-%Y'
# I only want the 1st column (dates) and 5 column (close), all rows
date = data[:,0][:,]
close = data[:,4][:,]
dates = [datetime.strptime(date, time_format)]
%plotting section
plt.plot_date(dates,close, '-')
plt.legend()
plt.show()
graph_data('stockhere')
Assuming the dates in the csv file are in the format '5-Jul-17', the proper format string to use is %d-%b-%y.
In [6]: datetime.strptime('5-Jul-17','%d-%m-%Y')
ValueError: time data '5-Jul-17' does not match format '%d-%m-%Y'
In [7]: datetime.strptime('5-Jul-17','%d-%b-%y')
Out[7]: datetime.datetime(2017, 7, 5, 0, 0)
See the Python documentation on strptime() behavior.