访问 Pandas DataFrame 元素内的列表
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Access a list within an element of a Pandas DataFrame
提问by Michael
I have a Pandas DataFramewhich 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 Ato the middle element of the list in Column B.
我有一个 Pandas DataFrame，其中一列中有一个整数列表。我想访问此列表中的各个元素。我找到了一种方法，通过使用tolist()并将其转换回DataFrame，但我想知道是否有更简单/更好的方法。在此示例中，我将 Column 添加A到Column中列表的中间元素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?
有一个更好的方法吗？
回答by breucopter
A bit more straightforward is:
更直接一点的是：
df['C'] = df['A'] + df['B'].apply(lambda x:x[1])
回答by Meilin He
You can also simply try the following:
您也可以简单地尝试以下操作：
df['C'] = df['A'] + df['B'].str[1]
Performance of this method:
该方法的性能：
%timeit df['C'] = df['A'] + df['B'].str[1]
#1000 loops, best of 3: 445 μs per loop
回答by Psidom
One option is to use the apply, which should be faster than creating a data frame out of it:
一种选择是使用apply，它应该比从中创建数据框更快：
df['C'] = df['A'] + df.apply(lambda row: row['B'][1], axis = 1)
Some speed test:
一些速度测试：
%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.
好的。稍微好一些。@breucopter 的答案是最快的。
