I'm using a pandas series and I want to find the index value that represents the quantile.
If I have:
np.random.seed(8) s = pd.Series(np.random.rand(6), ['a', 'b', 'c', 'd', 'e', 'f']) s a 0.873429 b 0.968541 c 0.869195 d 0.530856 e 0.232728 f 0.011399 dtype: float64 And do
s.quantile(.5) I get
0.70002511588475946 What I want to know is what is the index value of s that represents the point just before that quantile value. In this case I know the index value should be d.
If you set the interpolation argument to 'lower', 'higher', or 'nearest' then the problem can be solved a bit more simply as:
s[s == s.quantile(.5, interpolation='lower')] I'd guess this method is a fair bit faster than piRSquared's solution as well
Use sort_values, reverse the order, find all that are less than or equal to the quantile calculated, then find the idxmax.
(s.sort_values()[::-1] <= s.quantile(.5)).idxmax() Or:
(s.sort_values(ascending=False) <= s.quantile(.5)).idxmax() We can functionalize it:
def idxquantile(s, q=0.5, *args, **kwargs): qv = s.quantile(q, *args, **kwargs) return (s.sort_values()[::-1] <= qv).idxmax() idxquantile(s) 