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pythonpandasnumpyfrequencymode
Mykola Zotko
Mykola Zotko

Reputation: 17854

How to get the most frequent row in table

How to get the most frequent row in a DataFrame? For example, if I have the following table:

   col_1  col_2 col_3
0      1      1     A
1      1      0     A
2      0      1     A
3      1      1     A
4      1      0     B
5      1      0     C

Expected result:

   col_1  col_2 col_3
0      1      1     A

EDIT: I need the most frequent row (as one unit) and not the most frequent column value that can be calculated with the mode() method.

Upvotes: 15

Views: 729

Answers (5)

Mykola Zotko
Mykola Zotko

Reputation: 17854

In Pandas 1.1.0. is possible to use the method value_counts() to count unique rows in DataFrame:

df.value_counts()

Output:

col_1  col_2  col_3
1      1      A        2
       0      C        1
              B        1
              A        1
0      1      A        1

This method can be used to find the most frequent row:

df.value_counts().head(1).index.to_frame(index=False)

Output:

   col_1  col_2 col_3
0      1      1     A

Upvotes: 2

DDD1
DDD1

Reputation: 450

You can do this with groupby and size:

df = df.groupby(df.columns.tolist(),as_index=False).size()
result = df.iloc[[df["size"].idxmax()]].drop(["size"], axis=1)
result.reset_index(drop=True) #this is just to reset the index

Upvotes: 4

mathfux
mathfux

Reputation: 5949

npi_indexed library helps to perform some actions on 'groupby' type of problems with less script and similar performance as numpy. So this is alternative and pretty similar way to @Divakar's np.unique() based solution:

arr = df.values.astype(str)
idx = npi.multiplicity(arr)
output = df.iloc[[idx[c.argmax()]]]

Upvotes: 3

Divakar
Divakar

Reputation: 221614

With NumPy's np.unique -

In [92]: u,idx,c = np.unique(df.values.astype(str), axis=0, return_index=True, return_counts=True)

In [99]: df.iloc[[idx[c.argmax()]]]
Out[99]: 
   col_1  col_2 col_3
0      1      1     A

If you are looking for performance, convert the string column to numeric and then use np.unique -

a = np.c_[df.col_1, df.col_2, pd.factorize(df.col_3)[0]]
u,idx,c = np.unique(a, axis=0, return_index=True, return_counts=True)

Upvotes: 10

BENY
BENY

Reputation: 323316

Check groupby

df.groupby(df.columns.tolist()).size().sort_values().tail(1).reset_index().drop(0,1)
   col_1  col_2 col_3  
0      1      1     A

Upvotes: 12

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