Python Basic: Exercise-12 with Solution
Write a Python program to print the calendar of a given month and year.
Note: Use 'calendar' module.
Python calendar.month(theyear, themonth, w=0, l=0):
The function returns a month’s calendar in a multi-line string using the formatmonth() of the TextCalendar class.
'l' specifies the number of lines that each week will use.
import calendar y = int(input("Input the year : ")) m = int(input("Input the month : ")) print(calendar.month(y, m))
Input the year : 2017 Input the month : 04 April 2017 Mo Tu We Th Fr Sa Su 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Python Code Editor:
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Python: Tips of the Day
Find the index of an item in a list?
Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?
>>> ["foo", "bar", "baz"].index("bar") 1
Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:
list.index(x[, start[, end]])
Linear time-complexity in list length
An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:
>>> import timeit >>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000) 9.356267921015387 >>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000) 0.0004404920036904514
Only returns the index of the first match to its argument
A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.
>>> [1, 1].index(1) 0 >>> [i for i, e in enumerate([1, 2, 1]) if e == 1] [0, 2] >>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1) >>> next(g) 0 >>> next(g) 2
Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.
Throws if element not present in list
A call to index results in a ValueError if the item's not present.
>>> [1, 1].index(2) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: 2 is not in list
If the item might not be present in the list, you should either
- Check for it first with item in my_list (clean, readable approach), or
- Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)
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