Finding rows in a Pandas DataFrame with columns that violate a one-to-one mapping

Question

I have a DataFrame kinda like this:

| index | col_1 | col_2 |
| 0     | A     | 11    |
| 1     | B     | 12    |
| 2     | B     | 12    |
| 3     | C     | 13    |
| 4     | C     | 13    |
| 5     | C     | 14    |

where col_1 and col_2 may not always be one-to-one due to corrupt data.

How can I use Pandas to determine which rows have col_1 and col_2 entries that violate this one-to-one relationship?

In this case it would be the last three rows since C can either map to 13 or 14.

Answer 1

You could use a transform, counting the length of unique objects in each group. First look at the subset of just these columns, and then groupby a single column:

In [11]: g = df[['col1', 'col2']].groupby('col1')

In [12]: counts = g.transform(lambda x: len(x.unique()))

In [13]: counts
Out[13]:
  col2
0    1
1    1
2    1
3    2
4    2
5    2

The columns for the remaining columns (if not all)

In [14]: (counts == 1).all(axis=1)
Out[14]:
0     True
1     True
2     True
3    False
4    False
5    False
dtype: bool

Answer 2

I tested the g.transform(lambda x: len(x.unique())), works good but is slow especially when there are a lot of groups. The code below works much faster so I put it here.

df2 = pd.DataFrame(df[['col1', 'col2']].groupby(['col1','col2']).size(),columns=['count'])
df2.reset_index(inplace=True)
df3 = pd.DataFrame(df2.groupby('col1').size(), columns=['count'])
df4 = df3[df3['count']>1]
df_copy = df.copy()
df_copy.set_index('col1', inplace=True)
df_outlier = df_copy.ix[df4.index]

Answer 3

I would use a collections.Counter , because more than one instance of each item in a column violates the one-to-one mapping:

>>> import pandas
>>> import numpy
>>> import collections
>>> df = pandas.DataFrame(numpy.array([['a', 1],['b', 2], ['b', 3], ['c', 3]]))
>>> df
   0  1
0  a  1
1  b  2
2  b  3
3  c  3
>>> collections.Counter(df[0])
Counter({'b': 2, 'a': 1, 'c': 1})
>>> violations1 = [k for k, v in collections.Counter(df[0]).items() if v > 1]
>>> violations2 = [k for k, v in collections.Counter(df[1]).items() if v > 1]
>>> violations1
['b']
>>> violations2
['3']

Answer 4

Im super new to python but found a way to do it by gathering all the unique groupings into a list and filtering out the ones that were not uniquely mapped:

data = pd.DataFrame({'Col_1': ['A', 'B', 'B', 'C', 'C', 'C'], 'Col_2': [11,12,12,13,13,14]})
combos = []
for x, y in enumerate(range(len(data['Col_1']))):
    combo = '%s_%s' %(data['Col_1'][x], data['Col_2'][x])
    combos.append(combo)
data.index = data['Col_1']
for item in combos:
    if len([comb for comb in combos if item[2:] in comb[2:]]) != len([comb for comb in combos if item[0] in comb[0]]):
        data = data.drop(item[0])
data.reset_index(drop=True)