在Colab笔记本中,我想调用我在一个单独的python文件中编写的python函数。我该怎么做?
答案 0 :(得分:16)
修改:如果您要导入本地模块,则需要编辑sys.path
以指向该新目录。这是一个示例笔记本:
https://colab.research.google.com/notebook#fileId=1PtYW0hZit-B9y4PL978kV2ppJJPhjQua
原始回复: 当然,这是一个示例笔记本: https://colab.research.google.com/notebook#fileId=1KBrq8aAiy8vYIIUiTb5UHG9GKOdEMF3n
有两个单元格:第一个定义带有要导入函数的.py
文件。
%%writefile example.py
def f():
print 'This is a function defined in a Python source file.'
第二个单元格使用execfile
来评估笔记本的Python解释器中的.py
文件。
# Bring the file into the local Python environment.
execfile('example.py')
# Call the function defined in the file.
f()
答案 1 :(得分:3)
请尝试此功能将驱动器中的功能导入colab笔记本:
from google.colab import files
import zipfile, io, os
def upload_dir_file(case_f):
# author: yasser mustafa, 21 March 2018
# case_f = 0 for uploading one File or Package(.py) and case_f = 1 for uploading one Zipped Directory
uploaded = files.upload() # to upload a Full Directory, please Zip it first (use WinZip)
for fn in uploaded.keys():
name = fn #.encode('utf-8')
#print('\nfile after encode', name)
#name = io.BytesIO(uploaded[name])
if case_f == 0: # case of uploading 'One File only'
print('\n file name: ', name)
return name
else: # case of uploading a directory and its subdirectories and files
zfile = zipfile.ZipFile(name, 'r') # unzip the directory
zfile.extractall()
for d in zfile.namelist(): # d = directory
print('\n main directory name: ', d)
return d
print('Done!')
然后按照以下两个步骤操作: 1-如果您有一个名为(package_name.py)的文件,要将其上传到您的colab笔记本电话:
file_name = upload_dir_file(0)
2-然后,导入你的包裹:
import package_name
注意:您可以使用相同的功能: 1-上传文件(csv,excel,pdf,....):
file_name = upload_dir_file(0)
2-上传目录及其子目录和文件:
dir_name = upload_dir_file(1)
享受它!
答案 2 :(得分:0)
鲍勃·史密斯(Bob Smith)的答案无法在合作实验室中找到。 最简单的方法是:
exec(open(filename).read())
适用于所有版本。祝你好运!