熊猫矢量化而不是循环

Question

我有一个数据帧的路径。任务是使用类似datetime.fromtimestamp(os.path.getmtime('PATH_HERE'))成一个单独的列

import pandas as pd 
import numpy as np 
import os 


df1 = pd.DataFrame({'Path' : ['C:\\Path1' ,'C:\\Path2', 'C:\\Path3']}) 

#for a MVCE use the below commented out code. WARNING!!! This WILL Create directories on your machine. 
#for path in df1['Path']: 
# os.mkdir(r'PUT_YOUR_PATH_HERE\\' + os.path.basename(path)) 

我可以用下面的做任务得到的最后修改时间为文件夹，但它是一个缓慢的循环，如果我有很多文件夹：

for each_path in df1['Path']: 
    df1.loc[df1['Path'] == each_path, 'Last Modification Time'] = datetime.fromtimestamp(os.path.getmtime(each_path)) 

我该如何去引导这个过程来提高速度？ os.path.getmtime不能接受该系列。我在寻找类似：

df1['Last Modification Time'] = datetime.fromtimestamp(os.path.getmtime(df1['Path']))

Answer 1

我要去假设使用100条路径的3种方法。我认为方法3是优选的。

# Data initialisation: 
paths100 = ['a_whatever_path_here'] * 100 
df = pd.DataFrame(columns=['paths', 'time']) 
df['paths'] = paths100 


def fun1(): 
    # Naive for loop. High readability, slow. 
    for path in df['paths']: 
     mask = df['paths'] == path 
     df.loc[mask, 'time'] = datetime.fromtimestamp(os.path.getmtime(path)) 


def fun2(): 
    # Naive for loop optimised. Medium readability, medium speed. 
    for i, path in enumerate(df['paths']): 
     df.loc[i, 'time'] = datetime.fromtimestamp(os.path.getmtime(path)) 


def fun3(): 
    # List comprehension. High readability, high speed. 
    df['time'] = [datetime.fromtimestamp(os.path.getmtime(path)) for path in df['paths']] 


% timeit fun1() 
>>> 164 ms ± 2.03 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 

% timeit fun2() 
>>> 11.6 ms ± 67.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each) 

% timeit fun3() 
>>> 13.1 ns ± 0.0327 ns per loop (mean ± std. dev. of 7 runs, 100000000 loops each) 

Answer 2

可以使用GROUPBY transform（让你每组做昂贵的调用仅一次）：

g = df1.groupby("Path")["Path"] 
s = pd.to_datetime(g.transform(lambda x: os.path.getmtime(x.name))) 
df1["Last Modification Time"] = s # putting this on two lines so it looks nicer...