Removing Pandas duplicate values within rows, replace with NaNs, shifting NaNs to the end of rows

You can stack and then drop_duplicates that way. Then we need to pivot with the help of a cumcount level. The stack preserves the order the values appear in along the rows and the cumcount ensures that the NaN will appear in the end.

df1 = df.stack().reset_index().drop(columns='level_1').drop_duplicates()

df1['col'] = df1.groupby('level_0').cumcount()
df1 = (df1.pivot(index='level_0', columns='col', values=0)
          .rename_axis(index=None, columns=None))

   0  1    2    3
0  A  B    C    D
1  A  D    C  NaN
2  C  B  NaN  NaN
3  B  A  NaN  NaN

---

Timings

Assuming 4 columns, let's see how a bunch of these methods compare as the number of rows grow. The map and apply solutions have a good advantage when things are small, but they become a bit slower than the more involved stack + drop_duplicates + pivot solution as the DataFrame gets longer. Regardless, they all start to take a while for a large DataFrame.

import perfplot
import pandas as pd
import numpy as np

def stack(df):
    df1 = df.stack().reset_index().drop(columns='level_1').drop_duplicates()

    df1['col'] = df1.groupby('level_0').cumcount()
    df1 = (df1.pivot(index='level_0', columns='col', values=0)
              .rename_axis(index=None, columns=None))
    return df1

def apply_drop_dup(df):
    return pd.DataFrame.from_dict(df.apply(lambda x: x.drop_duplicates().tolist(),
                                           axis=1).to_dict(), orient='index')

def apply_unique(df):
    return pd.DataFrame(df.apply(pd.Series.unique, axis=1).tolist())


def list_map(df):
    return pd.DataFrame(list(map(pd.unique, df.values)))


perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(list('ABCD'), (n, 4)),
                                 columns=list('abcd')), 
    kernels=[
        lambda df: stack(df),
        lambda df: apply_drop_dup(df),
        lambda df: apply_unique(df),
        lambda df: list_map(df),
    ],
    labels=['stack', 'apply_drop_dup', 'apply_unique', 'list_map'],
    n_range=[2 ** k for k in range(18)],
    equality_check=lambda x,y: x.compare(y).empty,  
    xlabel='~len(df)'
)

---

Finally, if preserving the order in which the values originally appeared within each row is unimportant, you can use numpy. To de-duplicate you sort then check for differences. Then create an output array that shifts values to the right. Because this method will always return 4 columns, we require a dropna to match the other output in the case that every row has fewer than 4 unique values.

def with_numpy(df):
    arr = np.sort(df.to_numpy(), axis=1)
    r = np.roll(arr, 1, axis=1)
    r[:, 0] = np.NaN
    
    arr = np.where((arr != r), arr, np.NaN)
    
    # Move all NaN to the right. Credit @Divakar
    mask = pd.notnull(arr)
    justified_mask = np.flip(np.sort(mask, axis=1), 1)
    out = np.full(arr.shape, np.NaN, dtype=object) 
    out[justified_mask] = arr[mask]
    
    return pd.DataFrame(out, index=df.index).dropna(how='all', axis='columns')

with_numpy(df)
#   0  1    2    3
#0  A  B    C    D
#1  A  C    D  NaN
#2  B  C  NaN  NaN     # B/c this method sorts, B before C
#3  A  B  NaN  NaN

---

perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(list('ABCD'), (n, 4)),
                                 columns=list('abcd')), 
    kernels=[
        lambda df: stack(df),
        lambda df: with_numpy(df),
    ],
    labels=['stack', 'with_numpy'],
    n_range=[2 ** k for k in range(3, 22)],
    # Lazy check to deal with string/NaN and irrespective of sort order. 
    equality_check=lambda x, y: (np.sort(x.fillna('ZZ').to_numpy(), 1) 
                                 == np.sort(y.fillna('ZZ').to_numpy(), 1)).all(),
    xlabel='len(df)'
)

try something new

df = pd.DataFrame(list(map(pd.unique, df.values)))
Out[447]: 
   0  1     2     3
0  A  B     C     D
1  A  D     C  None
2  C  B  None  None
3  B  A  None  None

Use apply and construct a new dataframe by pd.DataFrame.from_dict with option orient='index'

df_final = pd.DataFrame.from_dict(df.apply(lambda x: x.drop_duplicates().tolist(),
                                               axis=1).to_dict(), orient='index')

Out[268]:
   0  1     2     3
0  A  B     C     D
1  A  D     C  None
2  C  B  None  None
3  B  A  None  None

Note: None practically is similar to NaN. If you want exact NaN. Just chain additional .fillna(np.nan)