I have a list that is filled with HTML elements. I also have a list filled with date/times, which is parallel to the HTML list.
How can I sort the HTML list based on the time/date list? The time/date is in a timestamp format.
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4 Answers
You can use zip.
timestamps, elements = zip(*sorted(zip(timestamps, elements)))
The result will be two tuples which you can convert to lists if you prefer.
4 Comments
Ben Blank
@Thomas — Don't feel bad!
sort() is in-place, while sorted() creates a new list, so sort() it typically what you want for big lists anyway. :-)
Zeno
Thanks. What about backwards?
canbax
it is very strange that the same function can zip and unzip values. This makes it hard to understand the function.
Epsilon
zip does not unzip, but the * unpacks the sorted zip object, which is then again zipped by zip.
Zip the two lists up into tuples, sort, then take the HTML back out of the tuple:
zipped = zip(timestamps, htmls)
zipped.sort()
sorted_htmls = [html for (timestamp, html) in zipped]
Comments
Enumerate will give you a list of (index,item) tuples for each item in the list. You can sort this using the index to read the sort key from the timestamps list. Then peel off the html elements:
sorted_elems = [elem for i,elem in sorted(enumerate(html_elements),
key=lambda x:timestamps[x[0]])]
What could be simpler?
Comments
You can use a third-party package called more_itertools:
import more_itertools
timestamps = ["08-11-17", "01-13-17", "11-22-17"]
elements = ["<p>", "<b>", "<a href>"]
more_itertools.sort_together([timestamps, elements])
# [('01-13-17', '08-11-17', '11-22-17'), ('<b>', '<p>', '<a href>')]
See more_itertools docs for more information.
