python：IOError：[Errno 2]使用sublimeREPL時沒有此類文件或目錄

Question

我已經將python文件和'g1.txt'放在同一目錄中。 當我不使用SublimeREPL時，代碼可以正確運行

def build_graph(file_name):
    new_file = open(file_name, 'r')
    n, m = [int(x) for x in new_file.readline().split()]

    graph = {}
    for line in new_file:
        # u, v, w is the tail, head and the weight of the a edge
        u, v, w = [int(x) for x in line.split()]
        graph[(u, v)] = w

    return n, graph

if __name__ == '__main__':
    print build_graph('g1.txt')

>>> >>> Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<string>", line 18, in <module>
  File "<string>", line 6, in build_graph
IOError: [Errno 2] No such file or directory: 'g1.txt'

Answer 1

嘗試這個：

 import os
 build_graph(os.path.join(os.path.dirname(__file__),"g1.txt"))

它將腳本的目錄附加到g1.txt

Answer 2

擴展這個答案 ，SublimeREPL不一定使用相同的工作目錄g1.txt是，你可以使用

import os
build_graph(os.path.join(os.path.dirname(__file__),"g1.txt"))

如先前建議的那樣，否則以下內容也將起作用：

if __name__ == '__main__':
    import os
    os.chdir(os.path.dirname(__file__))
    print build_graph('g1.txt')

---

只是一件小事，但您也不會關閉文件描述符。 您應該改用with open()格式：

def build_graph(file_name):
    with open(file_name, 'r') as new_file:
        n, m = [int(x) for x in new_file.readline().split()]

        graph = {}
        for line in new_file:
            # u, v, w is the tail, head and the weight of the a edge
            u, v, w = [int(x) for x in line.split()]
            graph[(u, v)] = w

    return n, graph

完成后，它將自動關閉文件描述符，因此您不必擔心手動關閉它。 將文件保持打開狀態通常不是一個好主意，特別是如果您正在寫文件時，因為程序結束時它們可能處於不確定狀態。