# Compute co-occurrence matrix by counting values in cells

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I have a dataframe like this

``df = pd.DataFrame({'a' : [1,1,0,0], 'b': [0,1,1,0], 'c': [0,0,1,1]}) ``

I want to get

``  a b c a 2 1 0 b 1 2 1 c 0 1 2 ``

where a,b,c are column names, and I get the values counting '1' in all columns when the filter is '1' in another column. For ample, when df.a == 1, we count a = 2, b =1, c = 0 etc

I made a loop to solve

``matrix = [] for name, values in df.iteritems():     matrix.append(pd.DataFrame( df.groupby(name, as_index=False).apply(lambda x: x[x == 1].count())).values.tolist()[1]) pd.DataFrame(matrix) ``

But I think that there is a simpler solution, isn't it?

You appear to want the matrix product, so leverage `DataFrame.dot`:

``df.T.dot(df)    a  b  c a  2  1  0 b  1  2  1 c  0  1  2 ``

Alternatively, if you want the same level of performance without the overhead of pandas, you could compute the product with `np.dot`:

``v = df.values pd.DataFrame(v.T.dot(v), index=df.columns, columns=df.columns) ``

Or, if you want to get cute,

``(lambda a, c: pd.DataFrame(a.T.dot(a), c, c))(df.values, df.columns) ``

``   a  b  c a  2  1  0 b  1  2  1 c  0  1  2 ``

—piRSquared