正如@Vivek Solanki所提到的，我也在协作仪表板的“文件”部分上传了我的文件。 只需要注意文件上传的位置。对我来说, train_data = pd.read_csv('/fileName.csv')有效。

上传

from google.colab import files
files.upload()

下载

files.download('filename')

目录列表

files.os.listdir()

快速，简单地从Dropbox导入:

!pip install dropbox
import dropbox
access_token = 'YOUR_ACCESS_TOKEN_HERE' # https://www.dropbox.com/developers/apps
dbx = dropbox.Dropbox(access_token)

# response = dbx.files_list_folder("")

metadata, res = dbx.files_download('/dataframe.pickle2')

with open('dataframe.pickle2', "wb") as f:
  f.write(res.content)

从谷歌。Colab导入驱动器

驱动器(' /内容/ drive’山)

进口熊猫作为pd dv = pd.read_csv(' /内容/传动/ MyDrive /戴安娜/卡索/ Data_Caso_Propuesto.csv”) dv.info ()

已解决，请在这里找到详细信息，并使用下面的功能: https://stackoverflow.com/questions/47212852/how-to-import-and-read-a-shelve-or-numpy-file-in-google-colaboratory/49467113#49467113

from google.colab import files
import zipfile, io, os

    def read_dir_file(case_f):
        # author: yasser mustafa, 21 March 2018  
        # case_f = 0 for uploading one File 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!')

在任何协作的左侧栏上都有一个称为“文件”的部分。 在那里上传文件并使用此路径

"/content/YourFileName.extension"

ex: pd read_csv(“/内容/ Forbes2015。csv”);