In-place sort_values in pandas what does it exactly mean?

Here an example. df1 will hold sorted dataframe and df will be intact import pandas as pd from datetime import datetime as dt df = pd.DataFrame(data=[22,22,3], index=[dt(2016, 11, 10, 0), dt(2016, 11, 10, 13), dt(2016, 11, 13, 5)], columns=[‘foo’]) df1 = df.sort_values(by=’foo’) print(df, df1) In the case below, df will hold sorted values import pandas … Read more

In-place modification of Python lists

I found this in the docs: https://docs.python.org/3/tutorial/controlflow.html#for Python’s for statement iterates over the items of any sequence (a list or a string), in the order that they appear in the sequence. If you need to modify the sequence you are iterating over while inside the loop (for example to duplicate selected items), it is recommended … Read more

Why aren’t Pandas operations in-place?

As evidenced here in the pandas documentation, “… In general we like to favor immutability where sensible.” The Pandas project is in the camp of preferring immutable (stateless) objects over mutable (objects with state) to guide programmers into creating more scalable / parallelizable data processing code. They are guiding the users by making the ‘inplace=False’ … Read more

Is Quicksort in-place or not? [duplicate]

Intro to Algorithms from MIT Press qualifies QuickSort as in-place – it sorts the elements within the array with at most a constant amount of them outside the array at any given time. At the end of the day, people will always have differing opinions (is Top-Down Memoization considered Dynamic Programming? Not to some “classical” … Read more

Pandas: peculiar performance drop for inplace rename after dropna

This is a copy of the explanation on github. There is no guarantee that an inplace operation is actually faster. Often they are actually the same operation that works on a copy, but the top-level reference is reassigned. The reason for the difference in performance in this case is as follows. The (df1-df2).dropna() call creates … Read more

Sort a part of a list in place

I’d write it this way: a[i:j] = sorted(a[i:j]) It is not in-place sort either, but fast enough for relatively small segments. Please note, that Python copies only object references, so the speed penalty won’t be that huge compared to a real in-place sort as one would expect.

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