Pandas return index and column name based on the item value

Question

I am trying to return a column name and index based on the item value. I have something like this:

So let's day I am trying to return index and column names of all values where value is > 0.75.

for date, row in df.iterrows():
    for item in row:
        if item > .75:
            print index, row

I wanted this to return "traffic and robbery". However this returns all the values. I did not find answer to this in documentation, online or here. Thank you in advance.

Answer 1

Using slightly different numbers (for no particular reason), you can stack to for a Series and then use boolean indexing:

In [11]: df.stack()
Out[11]:
assault  assault    1.00
         robbery    0.76
         traffic    0.60
robbery  assault    0.76
         robbery    1.00
         traffic    0.78
traffic  assault    0.68
         robbery    0.78
         traffic    1.00
dtype: float64

In [12]: s = df.stack()

In [13]: s[(s!=1) & (s>0.77)]
Out[13]:
robbery  traffic    0.78
traffic  robbery    0.78
dtype: float64

You can do a bit of numpy to remove the duplicates, one way* is to 0 those not in the upper diagonal with triu (unfortunately this doesn't return as a DataFrame :( ):

In [21]: np.triu(df, 1)
Out[21]:
array([[ 0.  ,  0.76,  0.6 ],
       [ 0.  ,  0.  ,  0.78],
       [ 0.  ,  0.  ,  0.  ]])

In [22]: s = pd.DataFrame(np.triu(df, 1), df.index, df.columns).stack() > 0.77

In [23]: s[s]
Out[23]:
robbery  traffic    True
dtype: bool

In [24]: s[s].index.tolist()
Out[24]: [('robbery', 'traffic')]

*I suspect there are more efficient ways...