Is there a pythonic way to do a contingency table in Pandas?

Question

Given a dataframe that looks like this:

            A   B      
2005-09-06  5  -2  
2005-09-07 -1   3  
2005-09-08  4   5 
2005-09-09 -8   2
2005-09-10 -2  -         


        

Answer 1

Here's a really useful page about the pandas crosstab function:

http://chrisalbon.com/python/pandas_crosstabs.html

So I think for what you'd like to do you should use

import pandas as pd
pd.crosstab(data['A']>0, data['B']>0)

Hope that helps!

Answer 2

Let us call your dataframe data. Try

a = data['A']>0
b = data['B']>0
data.groupby([a,b]).count() 

Answer 3

Probably easiest to just use the pandas function crosstab. Borrowing from Dyno Fu above:

import pandas as pd
from StringIO import StringIO
table = """dt          A   B
2005-09-06  5  -2
2005-09-07 -1   3
2005-09-08  4   5
2005-09-09 -8   2
2005-09-10 -2  -5
2005-09-11 -7   9
2005-09-12  2   8
2005-09-13  6  -5
2005-09-14  6  -5
"""
sio = StringIO(table)
df = pd.read_table(sio, sep=r"\s+", parse_dates=['dt'])
df.set_index("dt", inplace=True)

pd.crosstab(df.A > 0, df.B > 0)

Output:

B      False  True 
A                  
False      1      3
True       3      2

[2 rows x 2 columns]

Also the table is usable if you want to do a Fisher exact test with scipy.stats etc:

from scipy.stats import fisher_exact
tab = pd.crosstab(df.A > 0, df.B > 0)
fisher_exact(tab)

Answer 4

import pandas as pd
from StringIO import StringIO

table = """dt          A   B
2005-09-06  5  -2
2005-09-07 -1   3
2005-09-08  4   5
2005-09-09 -8   2
2005-09-10 -2  -5
2005-09-11 -7   9
2005-09-12  2   8
2005-09-13  6  -5
2005-09-14  6  -5
"""
sio = StringIO(table)
df = pd.read_table(sio, sep=r"\s+", parse_dates=['dt'])
df.set_index("dt", inplace=True)

a = df['A'] > 0
b = df['B'] > 0
df1 = df.groupby([a,b]).count()
print df1["A"].unstack()

output:

B      False  True
A
False      1      3
True       3      2

this is just lnanenok's answer and using unstack() to make it more readable. credit should go to lanenok.