Method/option for deduplicating index, a la drop_duplicates

[wesm]

see e.g. #2763

[patricktokeeffe]

The workarounds I've found for emulating DataFrame.drop_duplicates() on the index all use df.groupby(level=0) with .first() or .last(). Though easily rationalized, this is not intuitive. Also, IIUC this does not perform the operation in-place which could be a problem for very large data sets.

Maybe drop_duplicates could have a new boolean parameter which shortcuts like so

DataFrame.drop_duplicates(use_index=True) equals df.groupby(level=0).first()
DataFrame.drop_duplicates(use_index=True, take_last=True) equals df.groupby(level=0).last()

Of course the inplace parameter should still work. And if col is provided while use_index is True, then the index should be treated as just another column.

I'm wholly ignorant of what implementation barriers exist; this is just speculation about how it could be done.

[sinhrks]

Now we can do as below. Should document somewhere.

import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(10, 5), index=list('abcdeafagh'))
df[df.index.duplicated()]
#           0         1         2         3         4
# a -0.224246 -0.693000  0.424868 -0.384844  0.512322
# a -0.779163 -0.556674  0.102416 -0.761115  0.438872

df[~df.index.duplicated()]
#           0         1         2         3         4
# a  0.918487  0.438925  0.761677 -1.299548 -0.277456
# b  1.251101 -0.361700  3.210220 -0.778130 -0.351071
# c  0.768538  0.718754  0.565361  0.532387  0.797216
# d -0.147877  0.468789 -0.486800 -0.634428  0.502015
# e  0.331038  0.301599 -2.451796 -1.350511 -1.629918
# f -1.723805  0.714659 -0.219288  1.286036 -0.034969
# g -0.448491  0.874354  0.686063 -0.491885 -1.652858
# h -0.437542  1.518854  0.361197  2.423748 -0.425337

df[~df.index.duplicated(keep=False)]
#           0         1         2         3         4
# b  1.251101 -0.361700  3.210220 -0.778130 -0.351071
# c  0.768538  0.718754  0.565361  0.532387  0.797216
# d -0.147877  0.468789 -0.486800 -0.634428  0.502015
# e  0.331038  0.301599 -2.451796 -1.350511 -1.629918
# f -1.723805  0.714659 -0.219288  1.286036 -0.034969
# g -0.448491  0.874354  0.686063 -0.491885 -1.652858
# h -0.437542  1.518854  0.361197  2.423748 -0.425337

[bilderbuchi]

df[df.index.duplicated()]

Seeing how the index has a occuring 3 times, I have to say I would expect this statement to give 3 results.

[patricktokeeffe]

I hadn't noticed but totally agree. Labeling the two with higher index numbers as dups is arbitrary.

On August 11, 2015 1:05:51 PM PDT, Christoph Buchner notifications@github.com wrote:

df[df.index.duplicated()]

Seeing how the index has a occuring 3 times, I have to say I would
expect this statement to give 3 results.

---

Reply to this email directly or view it on GitHub:
#2825 (comment)

[sinhrks]

@bilderbuchi You can specify the behavior with keep option, see #10236

df[df.index.duplicated(keep=False)]
#           0         1         2         3         4
# a  0.791988 -0.127854  0.308921 -0.360801 -0.202838
# a -1.076777  1.027121  0.665178  0.115625  0.381496
# a  0.478709 -0.487886 -1.643373 -1.430937 -1.386072

df[~df.index.duplicated(keep=False)]
#           0         1         2         3         4
# b -1.773957  1.150821 -1.223845 -1.463176 -0.832168
# c -0.934962 -0.002800  0.102012  2.207798 -0.528325
# d  1.061423  1.362033  2.487682 -0.709634 -0.683065
# e  0.166217  0.156525 -0.535327  1.128811  0.120434
# f -0.542128  0.108480  0.572719  1.358334  0.159375
# g  0.264287 -0.188011  1.264308 -0.371283  0.089278
# h  0.059987 -0.604473 -1.010755  1.112758  0.865414

[patricktokeeffe]

Ah - that's subtle, but a very good compromise.

[bilderbuchi]

ah, interesting, thanks for the clarification.
yeah, that definitely should be documented somewhere imo.
Also, after my initial reading I read the discussion in #10236 - I also feel that a kwarg changing between string and bool feels reeeally weird (I'd have gone for 'none'), but I guess that ship has sailed. maybe also worth it to document that clearly.

[sinhrks]

Thanks to confirm. #10236 is included in v0.17 and not released yet, we can discuss if you have better alternatives.

[bilderbuchi]

Thanks. Well, as I said I'd have gone for a 'none' string instead of the boolean False - as a Python intermediate I find neither more pythonic than the other. I'm just a casual user of pandas, though, and I I'm sure that @jreback has more investment in pandas, so I guess that it fits better with the general pandas API or somesuch...

[jreback]

@bilderbuchi having a string and a Boolean possible in a keyword are no big deal

'none' however is very confusing

in any event @sinhrks out up s nice doc about how to do this