Pandas Dataframe: Show duplicate rows - with exact duplicates

You can use Nested ternary operator to iterate through your rows and get the duplicate values.

import numpy as np

df['Solution'] = df.apply(lambda row: row.col1 if row.col1 == row.col2 else row.col2 if row.col2 == row.col3 else row.col3 if row.col3 == row.col1 else np.NaN, axis=1)

which gives us the expected output

df
      col1    col2    col3 Solution
0    1-233   6-998   3-957      NaN
1   2-766g  2-766g     NaN   2-766g
2    6-455   5-955     NaN      NaN
3    4-356   7-236  3-602m      NaN
4    5-253   5-253   1-266    5-253
5    2-122   7-258   2-122    2-122
6    5-531  8-987t   7-834      NaN
7    8-345   7-567   8-345    8-345
8    1-505   1-505   2-858    1-505
9    3-127   6-876  7-984g      NaN
10   3-622     NaN     NaN      NaN

To keep the function readable and general, so it works for more or less than three cols, I'd just rely on writing a dedicated function that uses pandas built in functionality for finding duplicates, and applying that to the dataframe rows:

import numpy as np
import pandas as pd

df = pd.DataFrame({'col1':['1-233','2-766g','6-455','4-356','5-253','2-122','5-531','8- 345','1-505','3-127','3-622'],
'col2':['6-998','2-766g','5-955','7-236','5-253','7-258','8-987t','7-567','1-505','6-876','NaN'],
'col3':['3-957','NaN','NaN','3-602m','1-266','2-122','7-834','8-345','2-858','7-984g', 'NaN']})

def get_duplicate_value(row):
    """If row has duplicates, return that value, else NaN."""
    duplicate_locations = row.duplicated()
    if duplicate_locations.any():
        dup_index = duplicate_locations.idxmax()
        return row[dup_index]
    return np.NaN

df["solution"] = df.apply(get_duplicate_value, axis=1)

Check out the docs of pd.Dataframe.apply, pd.Series.duplicated, pd.Series.any and pd.Series.idxmax to figure out how this works exactly.

Output:

      col1    col2    col3 solution
0    1-233   6-998   3-957      NaN
1   2-766g  2-766g     NaN   2-766g
2    6-455   5-955     NaN      NaN
3    4-356   7-236  3-602m      NaN
4    5-253   5-253   1-266    5-253
5    2-122   7-258   2-122    2-122
6    5-531  8-987t   7-834      NaN
7   8- 345   7-567   8-345      NaN
8    1-505   1-505   2-858    1-505
9    3-127   6-876  7-984g      NaN
10   3-622     NaN     NaN      NaN

You could solve this by using Counter from the collections module.

import pandas as pd
import numpy as np
from collections import Counter

df = pd.DataFrame({'col1':['1-233','2-766g','6-455','4-356','5-253','2-122','5-531','8-345','1-505','3-127','3-622'],
'col2':['6-998','2-766g','5-955','7-236','5-253','7-258','8-987t','7-567','1-505','6-876','NaN'],
'col3':['3-957','NaN','NaN','3-602m','1-266','2-122','7-834','8-345','2-858','7-984g', 'NaN']})

def funct(row):
    if Counter(row).most_common(1)[0][1] >= 2:
        return Counter(row).most_common(1)[0][0]
    return np.nan

df['duplicate'] = df.iloc[:,:3].apply(funct, axis=1)

print(df) 
# N.B. I've changed the entry "8- 345" in `col1` to "8-345" 
# (the space appears to be a typo, judging from your prints).

      col1    col2    col3 duplicate
0    1-233   6-998   3-957       NaN
1   2-766g  2-766g     NaN    2-766g
2    6-455   5-955     NaN       NaN
3    4-356   7-236  3-602m       NaN
4    5-253   5-253   1-266     5-253
5    2-122   7-258   2-122     2-122
6    5-531  8-987t   7-834       NaN
7    8-345   7-567   8-345     8-345
8    1-505   1-505   2-858     1-505
9    3-127   6-876  7-984g       NaN
10   3-622     NaN     NaN       NaN

E.g. for row 1, Counter(row).most_common(1) will get us: Counter({'2-766g': 2, 'NaN': 1}). We want to check if the first (most common) value >= 2, so Counter(row).most_common(1)[0][1]. If so, the associated key will be your duplicate value. Else, we return np.nan.

Creating a second data frame and then inserting a new row into it with the duplicate values gives the required output.

column_names = ["col1", "col2", "col3", "duplicate_values"]
duplicated_df = pd.DataFrame(columns=column_names)
# code
for row in df.itertuples(index=True, name="Pandas"):
    seen = []
    ColumnIndex = int
    FOUND = False
    for j, value in enumerate(row):
        if value in seen:
            ColumnIndex = j
            FOUND = True
            break
        seen.append(value)
    DuplicateValue = "NaN"
    if FOUND:
        DuplicateValue = row[ColumnIndex]
    new_row = {
        "col1": row.col1,
        "col2": row.col2,
        "col3": row.col3,
        "duplicate_values": DuplicateValue,
    }
    # append row to the dataframe
    duplicated_df = duplicated_df.append(new_row, ignore_index=True)

print(duplicated_df)