Group duplicate column IDs in pandas dataframe

Question

Now there are a lot of similar questions but most of them answer how to delete the duplicate columns. However, I want to know how can I make a list of tuples where each tuple contains the column names of duplicate columns. I am assuming that each column has a unique name. Just to further illustrate my question:

df = pd.DataFrame({'A': [1, 2, 3, 4, 5],'B': [2, 4, 2, 1, 9],
                   'C': [1, 2, 3, 4, 5],'D': [2, 4, 2, 1, 9],
                   'E': [3, 4, 2, 1, 2],'F': [1, 1, 1, 1, 1]},
                   index = ['a1', 'a2', 'a3', 'a4', 'a5'])

then I want the output:

[('A', 'C'), ('B', 'D')]

And if you are feeling great today then also extend the same question to rows. How to get a list of tuples where each tuple contains duplicate rows.

Answer 1

Not using panda, just pure python :

data = {'A': [1, 2, 3, 4, 5],'B': [2, 4, 2, 1, 9],
        'C': [1, 2, 3, 4, 5],'D': [2, 4, 2, 1, 9],
        'E': [3, 4, 2, 1, 2],'F': [1, 1, 1, 1, 1]}
from collections import defaultdict

deduplicate = defaultdict(list)


for key, items in data.items():
    deduplicate[tuple(items)].append(key)  # cast to tuple because they are hashables but lists are not.

duplicates = list()
for vector, letters in deduplicate.items():
    if len(letters) > 1:
        duplicates.append(letters)

print(duplicates)

Using pandas :

import pandas

df = pandas.DataFrame(data)
duplicates = []

dedup2 = defaultdict(list)

for key in df.columns:
    dedup2[tuple(df[key])].append(key)

duplicates = list()
for vector, letters in dedup2.items():
    if len(letters) > 1:
        duplicates.append(letters)

print(duplicates)

Not really nice, but may be quicker since everything is done in one iteration over the data.

dedup2 = defaultdict(list)

duplicates = {}

for key in df.columns:
    astup = tuple(df[key])
    duplic = dedup2[astup] 
    duplic.append(key)
    if len(duplic) > 1:
        duplicates[astup] = duplic

duplicates = duplicates.values()
print(duplicates)

Answer 2

This is another approach that uses pure Python:

from operator import itemgetter
from itertools import groupby

def myfunc(df):
    # Convert the dataframe to a list of list including the column name
    zipped = zip(df.columns, df.values.T.tolist())
    # Sort the columns (so they can be grouped)
    zipped_sorted = sorted(zipped, key=itemgetter(1))
    # Placeholder for the result
    res = []
    res_append = res.append
    # Find duplicated columns using itertools.groupby
    for k, grp in groupby(zipped_sorted, itemgetter(1)):
        grp = list(grp)
        if len(grp) > 1:
            res_append(tuple(map(itemgetter(0), grp)))
    return res

I included some inline comments that illustrate how it works, but basically this just sorts the input so identical columns are adjacent and then it groups them.

I did some superficial timings using Divakars timing setup and got the following:

%timeit group_duplicate_cols(df)
391 ms ± 25.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit myfunc(df)
572 ms ± 4.36 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

So it seems like only 2 times slower than a NumPy approach, which is actually amazing.

Answer 3

Here's one NumPy approach -

def group_duplicate_cols(df):
    a = df.values
    sidx = np.lexsort(a)
    b = a[:,sidx]

    m = np.concatenate(([False], (b[:,1:] == b[:,:-1]).all(0), [False] ))
    idx = np.flatnonzero(m[1:] != m[:-1])
    C = df.columns[sidx].tolist()
    return [C[i:j] for i,j in zip(idx[::2],idx[1::2]+1)]

Sample runs -

In [100]: df
Out[100]: 
    A  B  C  D  E  F
a1  1  2  1  2  3  1
a2  2  4  2  4  4  1
a3  3  2  3  2  2  1
a4  4  1  4  1  1  1
a5  5  9  5  9  2  1

In [101]: group_duplicate_cols(df)
Out[101]: [['A', 'C'], ['B', 'D']]

# Let's add one more duplicate into group containing 'A'
In [102]: df.F = df.A

In [103]: group_duplicate_cols(df)
Out[103]: [['A', 'C', 'F'], ['B', 'D']]

Converting to do the same, but for rows(index), we just need to switch the operations along the other axis, like so -

def group_duplicate_rows(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])
    C = df.index[sidx].tolist()
    return [C[i:j] for i,j in zip(idx[::2],idx[1::2]+1)]

Sample run -

In [260]: df2
Out[260]: 
   a1  a2  a3  a4  a5
A   3   5   3   4   5
B   1   1   1   1   1
C   3   5   3   4   5
D   2   9   2   1   9
E   2   2   2   1   2
F   1   1   1   1   1

In [261]: group_duplicate_rows(df2)
Out[261]: [['B', 'F'], ['A', 'C']]

---

Benchmarking

Approaches -

# @John Galt's soln-1
from itertools import combinations
def combinations_app(df):
    return[x for x in combinations(df.columns, 2) if (df[x[0]] == df[x[-1]]).all()]

# @Abdou's soln
def pandas_groupby_app(df):
    return [tuple(d.index) for _,d in df.T.groupby(list(df.T.columns)) if len(d) > 1]                        

# @COLDSPEED's soln
def triu_app(df):
    c = df.columns.tolist()
    i, j = np.triu_indices(len(c), 1)
    x = [(c[_i], c[_j]) for _i, _j in zip(i, j) if (df[c[_i]] == df[c[_j]]).all()]
    return x

# @cmaher's soln
def lambda_set_app(df):
    return list(filter(lambda x: len(x) > 1, list(set([tuple([x for x in df.columns if all(df[x] == df[y])]) for y in df.columns]))))

Note : @John Galt's soln-2 wasn't included because the inputs being of size (8000,500) would blow up with the proposed broadcasting for that one.

Timings -

In [179]: # Setup inputs with sizes as mentioned in the question
     ...: df = pd.DataFrame(np.random.randint(0,10,(8000,500)))
     ...: df.columns = ['C'+str(i) for i in range(df.shape[1])]
     ...: idx0 = np.random.choice(df.shape[1], df.shape[1]//2,replace=0)
     ...: idx1 = np.random.choice(df.shape[1], df.shape[1]//2,replace=0)
     ...: df.iloc[:,idx0] = df.iloc[:,idx1].values
     ...: 

# @John Galt's soln-1
In [180]: %timeit combinations_app(df)
1 loops, best of 3: 24.6 s per loop

# @Abdou's soln
In [181]: %timeit pandas_groupby_app(df)
1 loops, best of 3: 3.81 s per loop

# @COLDSPEED's soln
In [182]: %timeit triu_app(df)
1 loops, best of 3: 25.5 s per loop

# @cmaher's soln
In [183]: %timeit lambda_set_app(df)
1 loops, best of 3: 27.1 s per loop

# Proposed in this post
In [184]: %timeit group_duplicate_cols(df)
10 loops, best of 3: 188 ms per loop

---

Super boost with NumPy's view functionality

Leveraging NumPy's view functionality that lets us view each group of elements as one dtype, we could gain further noticeable performance boost, like so -

def view1D(a): # a is array
    a = np.ascontiguousarray(a)
    void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
    return a.view(void_dt).ravel()

def group_duplicate_cols_v2(df):
    a = df.values
    sidx = view1D(a.T).argsort()
    b = a[:,sidx]

    m = np.concatenate(([False], (b[:,1:] == b[:,:-1]).all(0), [False] ))
    idx = np.flatnonzero(m[1:] != m[:-1])
    C = df.columns[sidx].tolist()
    return [C[i:j] for i,j in zip(idx[::2],idx[1::2]+1)]

Timings -

In [322]: %timeit group_duplicate_cols(df)
10 loops, best of 3: 185 ms per loop

In [323]: %timeit group_duplicate_cols_v2(df)
10 loops, best of 3: 69.3 ms per loop

Just crazy speedups!

Answer 4

Here's a single-liner

In [22]: from itertools import combinations

In [23]: [x for x in combinations(df.columns, 2) if (df[x[0]] == df[x[-1]]).all()]
Out[23]: [('A', 'C'), ('B', 'D')]

Alternatively, using NumPy broadcasting. Better, look at Divakar's solution

In [124]: cols = df.columns

In [125]: dftv = df.T.values

In [126]: cross = pd.DataFrame((dftv == dftv[:, None]).all(-1), cols, cols)

In [127]: cross
Out[127]:
       A      B      C      D      E      F
A   True  False   True  False  False  False
B  False   True  False   True  False  False
C   True  False   True  False  False  False
D  False   True  False   True  False  False
E  False  False  False  False   True  False
F  False  False  False  False  False   True

# Only take values from lower triangle
In [128]: s = cross.where(np.tri(*cross.shape, k=-1)).unstack()

In [129]: s[s == 1].index.tolist()
Out[129]: [('A', 'C'), ('B', 'D')]

Answer 5

Here's one more option using only comprehensions/built-ins:

filter(lambda x: len(x) > 1, list(set([tuple([x for x in df.columns if all(df[x] == df[y])]) for y in df.columns])))

Result:

[('A', 'C'), ('B', 'D')]

Answer 6

This should also do:

[tuple(d.index) for _,d in df.T.groupby(list(df.T.columns)) if len(d) > 1]

Yields:

# [('A', 'C'), ('B', 'D')]

Answer 7

Based on @John Galt one liner which is like this:

result_col = [x for x in combinations(df.columns, 2) if (df[x[0]] == df[x[-1]]).all()]

you can get the result_row as follows:

result_row = [x for x in combinations(df.T.columns,2) if (df.T[x[0]] == df.T[x[-1]]).all()]

using transpose (df.T)