我有一个pandas数据框,其列由列表组成。
目标是找到行中每个列表的最小值(以有效的方式)。
E.g。
import pandas as pd
df = pd.DataFrame(columns=['Lists', 'Min'])
df['Lists'] = [ [1,2,3], [4,5,6], [7,8,9] ]
print(df)
目标是Min
列:
Lists Min
0 [1, 2, 3] 1
1 [4, 5, 6] 4
2 [7, 8, 9] 7
提前谢谢你,
吉
答案 0 :(得分:4)
您可以apply
使用min
:
df['Min'] = df.Lists.apply(lambda x: min(x))
print (df)
Lists Min
0 [1, 2, 3] 1
1 [4, 5, 6] 4
2 [7, 8, 9] 7
感谢juanpa.arrivillaga的想法:
df['Min'] = [min(x) for x in df.Lists.tolist()]
print (df)
Lists Min
0 [1, 2, 3] 1
1 [4, 5, 6] 4
2 [7, 8, 9] 7
<强>计时强>:
##[300000 rows x 2 columns]
df = pd.concat([df]*100000).reset_index(drop=True)
In [144]: %timeit df['Min1'] = [min(x) for x in df.Lists.values.tolist()]
10 loops, best of 3: 137 ms per loop
In [145]: %timeit df['Min2'] = [min(x) for x in df.Lists.tolist()]
10 loops, best of 3: 142 ms per loop
In [146]: %timeit df['Min3'] = [min(x) for x in df.Lists]
10 loops, best of 3: 139 ms per loop
In [147]: %timeit df['Min4'] = df.Lists.apply(lambda x: min(x))
10 loops, best of 3: 170 ms per loop