按另一个数据帧中的列对熊猫数据帧进行排序 - 熊猫

Question

Let's say I have a Pandas DataFrame with two columns, like:

df = pd.DataFrame({'a': [1, 2, 3, 4], 'b': [100, 200, 300, 400]})
print(df)

---

   a    b
0  1  100
1  2  200
2  3  300
3  4  400

And let's say I also have a Pandas Series, like:

s = pd.Series([1, 3, 2, 4])
print(s)

---

0    1
1    3
2    2
3    4
dtype: int64

How can I sort the a column to become the same order as the s series, with the corresponding row values sorted together?

My desired output would be:

   a    b
0  1  100
1  3  300
2  2  200
3  4  400

Is there any way to achieve this?

Please check self-answer below.

Answer 1

What about:

(
    df.assign(s=s)
    .sort_values(by='s')
    .drop('s', axis=1)
)

Answer 2

I have ran into these issues quite often, so I just thought to share my solutions in Pandas.

Solutions:

Solution 1:

Using set_index to convert the a column to the index, then use reindex to change the order, then use rename_axis to change the index name back to a , then use reset_index to convert the a column from an index back to a column:

print(df.set_index('a').reindex(s).rename_axis('a').reset_index('a'))

Solution 2:

Using set_index to convert the a column to the index, then use loc to change the order, then use reset_index to convert the a column from an index back to a column:

print(df.set_index('a').loc[s].reset_index())

Solution 3:

Using iloc to index the rows in a different order, then use map to get that order that would fit the df to make it get sorted with the s series:

print(df.iloc[list(map(df['a'].tolist().index, s))])

Solution 4:

Using pd.DataFrame to create a new DataFrame object, then use sorted with a key argument to sort the DataFrame by the s series:

print(pd.DataFrame(sorted(df.values.tolist(), key=lambda x: s.tolist().index(x[0])), columns=df.columns))

Timings:

Timing with the below code:

import pandas as pd
from timeit import timeit
df = pd.DataFrame({'a': [1, 2, 3, 4], 'b': [100, 200, 300, 400]})
s = pd.Series([1, 3, 2, 4])
def u10_1():
    return df.set_index('a').reindex(s).rename_axis('a').reset_index('a')
def u10_2():
    return df.set_index('a').loc[s].reset_index()
def u10_3():
    return df.iloc[list(map(df['a'].tolist().index, s))]
def u10_4():
    return pd.DataFrame(sorted(df.values.tolist(), key=lambda x: s.tolist().index(x[0])), columns=df.columns)
print('u10_1:', timeit(u10_1, number=1000))
print('u10_2:', timeit(u10_2, number=1000))
print('u10_3:', timeit(u10_3, number=1000))
print('u10_4:', timeit(u10_4, number=1000))

Output:

u10_1: 3.012849470495621
u10_2: 3.072132612502147
u10_3: 0.7498072134665241
u10_4: 0.8109911930595484

@Allen has a pretty good answer too.