NumPy: DateTime Exercises, Practice, Solution

Last update on May 28 2022 13:25:56 (UTC/GMT +8 hours)

NumPy DateTime [7 exercises with solution]

[An editor is available at the bottom of the page to write and execute the scripts.]

1. Write a NumPy program to display all the dates for the month of March, 2017. Go to the editor
Expected Output:
March, 2017
['2017-03-01' '2017-03-02' '2017-03-03' '2017-03-04' '2017-03-05'
'2017-03-06' '2017-03-07' '2017-03-08' '2017-03-09' '2017-03-10'
'2017-03-11' '2017-03-12' '2017-03-13' '2017-03-14' '2017-03-15'
'2017-03-16' '2017-03-17' '2017-03-18' '2017-03-19' '2017-03-20'
'2017-03-21' '2017-03-22' '2017-03-23' '2017-03-24' '2017-03-25'
'2017-03-26' '2017-03-27' '2017-03-28' '2017-03-29' '2017-03-30'
'2017-03-31']
Click me to see the sample solution

2. Write a NumPy program to get the dates of yesterday, today and tomorrow. Go to the editor
Sample Output:
Yestraday: 2017-03-24
Today: 2017-03-25
Tomorrow: 2017-03-26
Click me to see the sample solution

3. Write a NumPy program to count the number of days of specific month. Go to the editor
Expected Output:
Number of days, February, 2016:
29 days
Number of days, February, 2017:
28 days
Number of days, February, 2018:
28 days
Click me to see the sample solution

4. Write a NumPy program to create 24 python datetime.datetime objects (single object for every hour), and then put it in a numpy array. Go to the editor
Expected Output:
[datetime.datetime(2000, 1, 1, 0, 0) datetime.datetime(2000, 1, 1, 1, 0)
datetime.datetime(2000, 1, 1, 2, 0) datetime.datetime(2000, 1, 1, 3, 0)
datetime.datetime(2000, 1, 1, 4, 0) datetime.datetime(2000, 1, 1, 5, 0)
datetime.datetime(2000, 1, 1, 6, 0) datetime.datetime(2000, 1, 1, 7, 0)
datetime.datetime(2000, 1, 1, 8, 0) datetime.datetime(2000, 1, 1, 9, 0)
datetime.datetime(2000, 1, 1, 10, 0) datetime.datetime(2000, 1, 1, 11, 0)
datetime.datetime(2000, 1, 1, 12, 0) datetime.datetime(2000, 1, 1, 13, 0)
datetime.datetime(2000, 1, 1, 14, 0) datetime.datetime(2000, 1, 1, 15, 0)
datetime.datetime(2000, 1, 1, 16, 0) datetime.datetime(2000, 1, 1, 17, 0)
datetime.datetime(2000, 1, 1, 18, 0) datetime.datetime(2000, 1, 1, 19, 0)
datetime.datetime(2000, 1, 1, 20, 0) datetime.datetime(2000, 1, 1, 21, 0)
datetime.datetime(2000, 1, 1, 22, 0) datetime.datetime(2000, 1, 1, 23, 0)]
Click me to see the sample solution

5. Write a NumPy program to find the first Monday in May 2017. Go to the editor
Expected Output:
First Monday in May 2017:
2017-05-01
Click me to see the sample solution

6. Write a NumPy program to find the number of weekdays in March 2017. Go to the editor
Note: "busday" default of Monday through Friday being valid days.
Sample Output:
Number of weekdays in March 2017:
23
Click me to see the sample solution

7. Write a NumPy program to convert numpy datetime64 to Timestamp. Go to the editor
Sample output:
Current date:
2017-04-01 08:01:12.722055
Timestamp:
1491033672.72
UTC from Timestamp:
2017-04-01 08:01:12.722055
Click me to see the sample solution

Python-Numpy Code Editor:

More to Come !

Do not submit any solution of the above exercises at here, if you want to contribute go to the appropriate exercise page.

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

Caveats follow

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.)

Ref: https://bit.ly/2ALwXwe