Find Indices Of Duplicate Rows In Pandas DataFrame

What is the pandas way of finding the indices of identical rows within a given DataFrame without iterating over individual rows? While it is possible to find all unique rows with u

Solution 1:

Use parameter duplicated with keep=False for all dupe rows and then groupby by all columns and convert index values to tuples, last convert output Series to list:

df = df[df.duplicated(keep=False)]

df = df.groupby(list(df)).apply(lambda x: tuple(x.index)).tolist()
print (df)
[(1, 6), (2, 4), (3, 5)]

If you want also see duplicate values:

df1 = (df.groupby(df.columns.tolist())
       .apply(lambda x: tuple(x.index))
       .reset_index(name='idx'))
print (df1)
   param_a  param_b  param_c     idx
0        0        0        0  (1, 6)
1        0        2        1  (2, 4)
2        2        1        1  (3, 5)

Solution 2:

Approach #1

Here's one vectorized approach inspired by this post-

def group_duplicate_index(df):
    a = df.values
    sidx = np.lexsort(a.T)
    b = a[sidx]

    m = np.concatenate(([False], (b[1:] == b[:-1]).all(1), [False] ))
    idx = np.flatnonzero(m[1:] != m[:-1])
    I = df.index[sidx].tolist()       
    return [I[i:j] for i,j in zip(idx[::2],idx[1::2]+1)]

Sample run -

In [42]: df
Out[42]: 
   param_a  param_b  param_c
1        0        0        0
2        0        2        1
3        2        1        1
4        0        2        1
5        2        1        1
6        0        0        0

In [43]: group_duplicate_index(df)
Out[43]: [[1, 6], [3, 5], [2, 4]]

Approach #2

For integer numbered dataframes, we could reduce each row to a scalar each and that lets us work with a 1D array, giving us a more performant one, like so -

def group_duplicate_index_v2(df):
    a = df.values
    s = (a.max()+1)**np.arange(df.shape[1])
    sidx = a.dot(s).argsort()
    b = a[sidx]

    m = np.concatenate(([False], (b[1:] == b[:-1]).all(1), [False] ))
    idx = np.flatnonzero(m[1:] != m[:-1])
    I = df.index[sidx].tolist() 
    return [I[i:j] for i,j in zip(idx[::2],idx[1::2]+1)]

---

Runtime test

Other approach(es) -

def groupby_app(df): # @jezrael's soln
    df = df[df.duplicated(keep=False)]
    df = df.groupby(df.columns.tolist()).apply(lambda x: tuple(x.index)).tolist()
    return df

Timings -

In [274]: df = pd.DataFrame(np.random.randint(0,10,(100000,3)))

In [275]: %timeit group_duplicate_index(df)
10 loops, best of 3: 36.1 ms per loop

In [276]: %timeit group_duplicate_index_v2(df)
100 loops, best of 3: 15 ms per loop

In [277]: %timeit groupby_app(df) # @jezrael's soln
10 loops, best of 3: 25.9 ms per loop