比較Python中字符串元素的最快方法

Question

我正在尋找比較python中字符串元素的最快方法。

import os, glob, numpy as np

with open ('fname.txt','r') as fi:   ##This infile contains 9 thousands of string elements
    all_list = fi.read().splitlines()

existing_list = glob.glob('*jpg') ##This contains 5 thousands elements
existing_list = [os.path.basename(f) for f in existing_list]

remaining_list = [f for f in all_list if f not in existing_list]
for i in remaining list:
    print i

如何在Numpy中執行？

all_list = np.array(all_list)
existing_list = np.array(existing_list)
remaining_list = ???

Answer 1

如果使用集合，則可以在不使用numpy的情況下對其進行優化：

existing_set = {os.path.basename(f) for f in existing_list}  # set comprehension, python2.7+
# alternatively:  set(os.path.basename(f) for f in existing_list)

remaining_list = [f for f in all_list if f not in existing_set]

我懷疑即使您想出一種方法也可以通過使用numpy來獲得很多性能...