I'm having a weird situation, where pd.describe is giving me percentile markers that disagree with scipy.stats percentileofscore, because of NaNs, I think.
My df is:
f_recommend
0 3.857143
1 4.500000
2 4.458333
3 NaN
4 3.600000
5 NaN
6 4.285714
7 3.587065
8 4.200000
9 NaN
When I run df.describe(percentiles=[.25, .5, .75]) I get:
f_recommend
count 7.000000
mean 4.069751
std 0.386990
min 3.587065
25% 3.728571
50% 4.200000
75% 4.372024
max 4.500000
I get the same values when I run with NaN removed.
When I want to look up a specific value, however, when I run scipy.stats.percentileofscore(df['f_recommend'], 3.61, kind = 'mean') I get: 28th percentile with NaN and 20th without.
Any thoughts to explain this discrepancy?
ETA:
I don't believe that the problem is that we're calculating percentiles differently. Because that only matters when you're calculating percentiles of the same 2 numbers in different ways. But here, describe gives 25 percentile as 3.72. So there is absolutely no way that 3.61 can be 28th percentile. None of the formulas should give that.
In particular, when I use describe on the 6 values without NaN, I get the same values, so that's ignoring NaN, which is fine. But when I run percentile of score without the NaN I get a number that doesn't match.
ETA 2:
Simpler example:
In [48]: d = pd.DataFrame([1,2,3,4,5,6,7])
In [49]: d.describe()
Out[49]:
0
count 7.000000
mean 4.000000
std 2.160247
min 1.000000
25% 2.500000
50% 4.000000
75% 5.500000
max 7.000000
In [50]: sp.stats.percentileofscore(d[0], 2.1, kind = 'mean')
Out[50]: 28.571428571428573
the "kind" argument doesn't matter because 2.1 is unique.
scipy.stats.percentileofscore does not ignore nan, nor does it check for the value and handle it in some special way. It is just another floating point value in your data. This means the behavior of percentileofscore with data containing nan is undefined, because of the behavior of nan in comparisons:
In [44]: np.nan > 0
Out[44]: False
In [45]: np.nan < 0
Out[45]: False
In [46]: np.nan == 0
Out[46]: False
In [47]: np.nan == np.nan
Out[47]: False
Those results are all correct--that is how nan is supposed to behave. But that means, in order to know how percentileofscore handles nan, you have to know how the code does comparisons. And that is an implementation detail that you shouldn't have to know, and that you can't rely on to be the same in future versions of scipy.
If you investigate the behavior of percentfileofscore, you'll find that it behaves as if nan was infinite. For example, if you replace nan with a value larger than any other value in the input, you'll get the same results:
In [53]: percentileofscore([10, 20, 25, 30, np.nan, np.nan], 18)
Out[53]: 16.666666666666664
In [54]: percentileofscore([10, 20, 25, 30, 999, 999], 18)
Out[54]: 16.666666666666664
Unfortunately, you can't rely on this behavior. If the implementation changes in the future, nan might end up behaving like negative infinity, or have some other unspecified behavior.
The solution to this "problem" is simple: don't give percentileofscore any nan values. You'll have to clean up your data first. Note that this can be as simple as:
result = percentileofscore(a[~np.isnan(a)], score)
the answer is very simple.
There is no universally accepted formula for computing percentiles, in particular when your data contains ties or when it cannot be perfectly broken down in equal-size buckets.
For instance, have a look at the documentation in R. There are more than seven types of formulas! https://stat.ethz.ch/R-manual/R-devel/library/stats/html/quantile.html
At the end, it comes down to understanding which formula is used and whether the differences are big enough to be a problem in your case.
