问题描述
我正在尝试打开一个 CSV 文件，但由于某种原因 python 无法找到它.
I am trying to open a CSV file but for some reason python cannot locate it.
这是我的代码(这只是一个简单的代码，但我无法解决问题):
Here is my code (it's just a simple code but I cannot solve the problem):
import csv
with open('address.csv','r') as f:
reader = csv.reader(f)
for row in reader:
print row
推荐答案
当你打开一个名为 address.csv 的文件时，你是在告诉 open()您的文件在当前工作目录中的功能.这称为相对路径.
When you open a file with the name address.csv, you are telling the open() function that your file is in the current working directory. This is called a relative path.
为了让您了解这意味着什么，请将其添加到您的代码中:
To give you an idea of what that means, add this to your code:
import os
cwd = os.getcwd() # Get the current working directory (cwd)
files = os.listdir(cwd) # Get all the files in that directory
print("Files in %r: %s" % (cwd, files))
这将打印当前工作目录以及其中的所有文件.
That will print the current working directory along with all the files in it.
告诉 open() 函数的另一种方法是使用绝对路径，例如:
Another way to tell the open() function where your file is located is by using an absolute path, e.g.:
f = open("/Users/foo/address.csv")
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