串联 python 中的两个文件
Concatenation two files in python
我有以下读取文件的函数,然后将它们存储在变量中:
def browseFiles():
filename = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename)
with open(filename) as fp:
firstfile = fp.read()
def browseFiles1():
filename1 = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename1)
with open(filename1) as fp:
secondfile = fp.read()
我想将第一个文件和第二个文件连接在一起,然后生成第三个文件。所以,我用了:
firstfile += "\n"
firstfile += secondfile
with open ('thirdfile.docx', 'w') as fp:
fp.write(firstfile)
我的问题是如何访问每个函数中的变量 firstfile 和 secondfile 并使用它们生成第三个文件?
您可以 return
来自这两个函数的文件内容,这就是您从“主函数”访问这两个函数的方式。
您必须 return 来自 2 个函数的 firstfile
和 secondfile
,将它们存储在变量中,然后使用 pd.concat
函数。
def browseFiles():
filename = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename)
with open(filename) as fp:
firstfile = fp.read()
return firstfile
def browseFiles1():
filename1 = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename1)
with open(filename1) as fp:
secondfile = fp.read()
return secondfile
firstfile = browseFiles()
secondfile = browseFiles1()
thirdfile = pd.concat([firstfile, secondfile])
这是 concat 文档的link。
干杯!
我有以下读取文件的函数,然后将它们存储在变量中:
def browseFiles():
filename = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename)
with open(filename) as fp:
firstfile = fp.read()
def browseFiles1():
filename1 = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename1)
with open(filename1) as fp:
secondfile = fp.read()
我想将第一个文件和第二个文件连接在一起,然后生成第三个文件。所以,我用了:
firstfile += "\n"
firstfile += secondfile
with open ('thirdfile.docx', 'w') as fp:
fp.write(firstfile)
我的问题是如何访问每个函数中的变量 firstfile 和 secondfile 并使用它们生成第三个文件?
您可以 return
来自这两个函数的文件内容,这就是您从“主函数”访问这两个函数的方式。
您必须 return 来自 2 个函数的 firstfile
和 secondfile
,将它们存储在变量中,然后使用 pd.concat
函数。
def browseFiles():
filename = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename)
with open(filename) as fp:
firstfile = fp.read()
return firstfile
def browseFiles1():
filename1 = filedialog.askopenfilename(initialdir = "/home",
title = "Select a File",
filetypes = (("Text files",
"*.docx*"),
("all files",
"*.*")))
label_file_explorer.configure(text="File Opened: "+ filename1)
with open(filename1) as fp:
secondfile = fp.read()
return secondfile
firstfile = browseFiles()
secondfile = browseFiles1()
thirdfile = pd.concat([firstfile, secondfile])
这是 concat 文档的link。
干杯!