Python: Check whether a given month and year contains a Monday 13th
Python Basic - 1: Exercise-130 with Solution
Write a Python program to check whether a given month and year contains a Monday 13th.
Sample Solution-1:
Python Code:
from datetime import date
def test(month, year):
return str(date(year,month,13).strftime("%A")=='Monday')
month = 11;
year = 2022;
print("Month No.: ", month, " Year: ", year);
print("Check whether the said month and year contains a Monday 13th.: " + test(month, year));
month = 6;
year = 2022;
print("\nMonth No.: ", month, " Year: ", year);
print("Check whether the said month and year contains a Monday 13th.: " + test(month, year));
Sample Output:
Month No.: 11 Year: 2022 Check whether the said month and year contains a Monday 13th.: False Month No.: 6 Year: 2022 Check whether the said month and year contains a Monday 13th.: True
Flowchart:

Sample Solution-2:
Python Code:
import calendar
def test(month, year):
return str(calendar.weekday(year, month, 13) == 0)
month = 11;
year = 2022;
print("Month No.: ", month, " Year: ", year);
print("Check whether the said month and year contains a Monday 13th.: " + test(month, year));
month = 6;
year = 2022;
print("\nMonth No.: ", month, " Year: ", year);
print("Check whether the said month and year contains a Monday 13th.: " + test(month, year));
Sample Output:
Month No.: 11 Year: 2022 Check whether the said month and year contains a Monday 13th.: False Month No.: 6 Year: 2022 Check whether the said month and year contains a Monday 13th.: True
Flowchart:

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Python: Tips of the Day
Iterating over dictionaries using 'for' loops:
I am a bit puzzled by the following code: d = {'x': 1, 'y': 2, 'z': 3} for key in d: print key, 'corresponds to', d[key] What I don't understand is the key portion. How does Python recognize ...
key is just a variable name.
for key in d:
For Python 3.x:
for key, value in d.items():
For Python 2.x:
for key, value in d.iteritems():
To test for yourself, change the word key to poop.
In Python 3.x, iteritems() was replaced with simply items(), which returns a set-like view backed by the dict, like iteritems() but even better. This is also available in 2.7 as viewitems().
The operation items() will work for both 2 and 3, but in 2 it will return a list of the dictionary's (key, value) pairs, which will not reflect changes to the dict that happen after the items() call. If you want the 2.x behavior in 3.x, you can call list(d.items()).
Ref: https://bit.ly/37dm0Qo
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