Why is pandas.Series.std() different from numpy.std()?

Question

This is what I am trying to explain:

>>> pd.Series([7,20,22,22]).std()
7.2284161474004804
>>> np.std([7,20,22,22])
6.2599920127744575

Answer: this is explained by Bessel's correction, N-1 instead of N in the denominator of the standard deviation formula. I wish Pandas used the same convention as numpy.

---

There is a related discussion here, but their suggestions do not work either.

I have data about many different restaurants. Here is my dataframe (imagine more than one restaurant, but the effect is reproduced with just one):

>>> df
restaurant_id  price
id                      
1           10407      7
3           10407     20
6           10407     22
13          10407     22

Question: r.mi.groupby('restaurant_id')['price'].mean() returns price means for each restaurant. I want to get the standard deviations. However, r.mi.groupby('restaurant_id')['price'].std() returns wrong values.

As you can see, for simplicity I have extracted just one restaurant with four items. I want to find the standard deviation of the price. Just to make sure:

>>> np.mean([7,20,22,22])
17.75
>>> np.std([7,20,22,22])
6.2599920127744575

We can get the same (correct) values with

>>> np.mean(df)
restaurant_id    10407.00
price               17.75
dtype: float64
>>> np.std(df)
restaurant_id    0.000000
price            6.259992
dtype: float64

(Of course, disregard the mean restaurant id.) Obviously, np.std(df) is not a solution when I have more than one restaurant. So I am using groupby.

>>> df.groupby('restaurant_id').agg('std')
                  price
restaurant_id          
10407          7.228416

What?! 7.228416 is not 6.259992.

Let's try again.

>>> df.groupby('restaurant_id').std()

Same thing.

>>> df.groupby('restaurant_id')['price'].std()

Same thing.

>>> df.groupby('restaurant_id').apply(lambda x: x.std())

Same thing.

However, this works:

for id, group in df.groupby('restaurant_id'):
  print id, np.std(group['price'])

Question: is there a proper way to aggregate the dataframe, so I will get a new time series with the standard deviations for each restaurant?

Answer 1

I see. Pandas is using Bessel's correction by default -- that is, the standard deviation formula with N-1 instead of N in the denominator. As behzad.nouri has pointed out in the comments,

pd.Series([7,20,22,22]).std(ddof=0)==np.std([7,20,22,22])