Is df.reindex(columns=reversed(df.columns)) the fastest way to reverse a pandas.DataFrame by column?
Asked
Active
Viewed 2,409 times
2
-
1Related: [What is the Big O Complexity of Reversing the Order of Columns in Pandas DataFrame?](https://stackoverflow.com/q/51486063/9209546) – jpp Jan 21 '19 at 15:56
2 Answers
3
One idea - use DataFrame.iloc with indexing:
df = df.iloc[:, ::-1]
Performance:
np.random.seed(234)
df = pd.DataFrame(np.random.randint(10, size=(3, 10000))).rename(columns=str)
#print (df)
In [225]: %timeit df.reindex(columns=reversed(df.columns))
7.32 ms ± 166 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [226]: %timeit df.iloc[:, ::-1]
132 µs ± 6.02 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
#vdkotian solution
In [227]: %timeit df[df.columns[::-1]]
2.84 ms ± 60.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
jezrael
- 822,522
- 95
- 1,334
- 1,252
0
Just tried on sample dataset.
sepal_length sepal_width petal_length petal_width species
0 5.1 3.5 1.4 0.2 setosa
1 4.9 3.0 1.4 0.2 setosa
2 4.7 3.2 1.3 0.2 setosa
3 4.6 3.1 1.5 0.2 setosa
4 5.0 3.6 1.4 0.2 setosa
5 5.4 3.9 1.7 0.4 setosa
6 4.6 3.4 1.4 0.3 setosa
7 5.0 3.4 1.5 0.2 setosa
8 4.4 2.9 1.4 0.2 setosa
9 4.9 3.1 1.5 0.1 setosa
>>> df[df.columns[::-1]]
species petal_width petal_length sepal_width sepal_length
0 setosa 0.2 1.4 3.5 5.1
1 setosa 0.2 1.4 3.0 4.9
2 setosa 0.2 1.3 3.2 4.7
3 setosa 0.2 1.5 3.1 4.6
4 setosa 0.2 1.4 3.6 5.0
5 setosa 0.4 1.7 3.9 5.4
6 setosa 0.3 1.4 3.4 4.6
7 setosa 0.2 1.5 3.4 5.0
8 setosa 0.2 1.4 2.9 4.4
9 setosa 0.1 1.5 3.1 4.9
df.column[::-1] reverses the order of the dataframe in sequence of column and df[..] re-indexes it.
A more succinct way to achieve the same thing is with the iloc indexer:
df.iloc[:, ::-1]
The first : means "take all rows", the ::-1 means step backwards through the columns.
for reference see link https://stackoverflow.com/questions/27817718/reverse-dataframe-column-order
vdkotian
- 539
- 6
- 13