问题
I have a Pandas DataFrame which has a list of integers inside one of the columns. I'd like to access the individual elements within this list. I've found a way to do it by using tolist() and turning it back into a DataFrame, but I am wondering if there is a simpler/better way. In this example, I add Column A to the middle element of the list in Column B.
import pandas as pd
df = pd.DataFrame({'A' : (1,2,3), 'B': ([0,1,2],[3,4,5,],[6,7,8])})
df['C'] = df['A'] + pd.DataFrame(df['B'].tolist())[1]
df
Is there a better way to do this?
回答1:
A bit more straightforward is:
df['C'] = df['A'] + df['B'].apply(lambda x:x[1])
回答2:
One option is to use the apply, which should be faster than creating a data frame out of it:
df['C'] = df['A'] + df.apply(lambda row: row['B'][1], axis = 1)
Some speed test:
%timeit df['C'] = df['A'] + pd.DataFrame(df['B'].tolist())[1]
# 1000 loops, best of 3: 567 µs per loop
%timeit df['C'] = df['A'] + df.apply(lambda row: row['B'][1], axis = 1)
# 1000 loops, best of 3: 406 µs per loop
%timeit df['C'] = df['A'] + df['B'].apply(lambda x:x[1])
# 1000 loops, best of 3: 250 µs per loop
OK. Slightly better. @breucopter's answer is the fastest.
回答3:
You can also simply try the following:
df['C'] = df['A'] + df['B'].str[1]
Performance of this method:
%timeit df['C'] = df['A'] + df['B'].str[1]
#1000 loops, best of 3: 445 µs per loop
来源：https://stackoverflow.com/questions/38088419/access-a-list-within-an-element-of-a-pandas-dataframe