python路径拼接os.path.join()函数

os.path.join()函数用于路径拼接文件路径。

拼接如下目录：

代码： 

 # -*- encoding:utf-8 -*-
import sys
import os
if __name__ == '__main__':
    if len(sys.argv) != 2:
        print('need input: task name , but got {0} '.format(len(sys.argv)))
        sys.exit(-1)
    task_name = sys.argv[1]
    congig_new = task_name + '_config.json'
    config_path = os.path.join('./', task_name, congig_new)
    print('config_path: ', config_path)
    model_path = os.path.join('./', task_name + '/model')
    data_path = os.path.join('./', task_name + '/data')
    print('model_path: ', model_path)
    print('data_path: ', data_path)

输出：

(torch_20201223) localhost:bert_train wang$ python test_demo.py fenleizhuanlan
config_path:  ./fenleizhuanlan/fenleizhuanlan_config.json
model_path:  ./fenleizhuanlan/model
data_path:  ./fenleizhuanlan/data

注意拼接目录时 "/"的使用，如果有“/”，则跟前面的目录用“+”拼接，否则，用逗号。也就是说os.path.join()的每一个参数都是一个目录名字，或者文件名字。

看一个跟上述代码达到同样功能的代码：

 # -*- encoding:utf-8 -*-
import sys
import os
if __name__ == '__main__':
    if len(sys.argv) != 2:
        print('need input: task name , but got {0} '.format(len(sys.argv)))
        sys.exit(-1)
    task_name = sys.argv[1]
    congig_new = task_name + '_config.json'
    config_path = os.path.join('./', task_name, congig_new)
    print('config_path: ', config_path)
    model_path = os.path.join('./', task_name, 'model')
    data_path = os.path.join('./', task_name, 'data')
    print('model_path: ', model_path)
    print('data_path: ', data_path)