T-test in Pandas

Question:

If I want to calculate the mean of two categories in Pandas, I can do it like this:

data = {'Category': ['cat2','cat1','cat2','cat1','cat2','cat1','cat2','cat1','cat1','cat1','cat2'],
        'values': [1,2,3,1,2,3,1,2,3,5,1]}
my_data = DataFrame(data)
my_data.groupby('Category').mean()

Category:     values:   
cat1     2.666667
cat2     1.600000

I have a lot of data formatted this way, and now I need to do a T-test to see if the mean of cat1 and cat2 are statistically different. How can I do that?

Asked By: hirolau

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Answers:

it depends what sort of t-test you want to do (one sided or two sided dependent or independent) but it should be as simple as:

from scipy.stats import ttest_ind

cat1 = my_data[my_data['Category']=='cat1']
cat2 = my_data[my_data['Category']=='cat2']

ttest_ind(cat1['values'], cat2['values'])
>>> (1.4927289925706944, 0.16970867501294376)

it returns a tuple with the t-statistic & the p-value

see here for other t-tests http://docs.scipy.org/doc/scipy/reference/stats.html

Answered By: Gonzalo

EDIT: I had not realized this was about the data format. You could use

import pandas as pd
import scipy
two_data = pd.DataFrame(data, index=data['Category'])

Then accessing the categories is as simple as

scipy.stats.ttest_ind(two_data.loc['cat'], two_data.loc['cat2'], equal_var=False)

The loc operator accesses rows by label.


As @G Garcia said

one sided or two sided dependent or independent

If you have two independent samples but you do not know that they have equal variance, you can use Welch’s t-test. It is as simple as

scipy.stats.ttest_ind(cat1['values'], cat2['values'], equal_var=False)

For reasons to prefer Welch’s test, see https://stats.stackexchange.com/questions/305/when-conducting-a-t-test-why-would-one-prefer-to-assume-or-test-for-equal-vari.

For two dependent samples, you can use

scipy.stats.ttest_rel(cat1['values'], cat2['values'])
Answered By: serv-inc

I simplify the code a little bit.

from scipy.stats import ttest_ind
ttest_ind(*my_data.groupby('Category')['value'].apply(lambda x:x.values))
Answered By: Ningrong Ye