/anaconda3/lib/python3.5/zipfile.py error when attempting to write to an Excel file

patelbg2001 used Ask the Experts™
Hi, I'm running a python script using Spyder (Python 3.5) on OSX 10.13.6 and keep on running into file not found errors when attempting to copy data into a Microsoft Excel file. It looks like a permissions issue, but I don't know how to resolve. Any help would be much appreciated,

import urllib.request
import pandas as pd
import sys
import os

sys.path.append('/Users/bhav\ 1/Documents/keepbak2/python')

from html_table_parser import HTMLTableParser

target = 'weblink'

#get website config
req = urllib.request.Request(url=target)
f = urllib.request.urlopen(req)
xhtml = f.read().decode('utf-8')

#instantiate he parser and feed it
p = HTMLTableParser()

#write the table into python list

cwd = os.getcwd()  # Get the current working directory (cwd)

Table_List = p.tables

# Use the Pandas from_dict method to read the tables into
# a Pandas data frame

df = pd.DataFrame.from_dict(Table_List[0])

#CountryDied_Stats.to_excel('/Users/bhav\ 1/Documents/keepbak2/python/countrydiedtest.xlsx',
#                           na_rep='',
#                           float_format=None, columns=None,
#                           header=True, index=True,
#                           index_label=None, startrow=0,
#                           startcol=0, engine=None, merge_cells=True,
#                           encoding=None, inf_rep='inf', verbose=True)

#xlfile = 'countrydiedtest.xlsx'
xlfile = '/Users/bhav\ 1/Documents/keepbak2/python/countrydiedtest.xlsx '

cwd = os.getcwd()  # Get the current working directory (cwd)
files = os.listdir(cwd)  # Get all the files in that directory
print("Files in '%s': %s" % (cwd, files))

# open file for reading
fn = input(xlfile)
if os.path.exists(fn):
#    fh = open(fn, "r")
    print("row 56 open(fn, r)")
#    fh = open(fn, "w")
    print(" row 59 open(fn, w)")

# Create a Pandas Excel writer using XlsxWriter as the engine.
writer = pd.ExcelWriter(fn, engine='xlsxwriter')

# Convert the dataframe to an XlsxWriter Excel object.
df.to_excel(writer, sheet_name='Sheet1')

# Close the Pandas Excel writer and output the Excel file.

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runfile('/Users/bhav 1/Documents/keepbak2/python/html-table-parser/scrape1.py', wdir='/Users/bhav 1/Documents/keepbak2/python/html-table-parser')
Reloaded modules: html_table_parser.parser, html_table_parser
/Users/bhav 1/Documents/keepbak2/python/html-table-parser
Files in '/Users/bhav 1/Documents/keepbak2/python/html-table-parser': ['countrydiedtest.xlsx', 'LICENSE', 'README.md', 'scrape1.py', '.gitignore', '.python-version', 'html_table_parser', 'example_of_usage.py', 'html_table_converter']

/Users/bhav 1/Documents/keepbak2/python/countrydiedtest.xlsx
 row 59 open(fn, w)
Traceback (most recent call last):

  File "<ipython-input-3-e281686b8449>", line 1, in <module>
    runfile('/Users/bhav 1/Documents/keepbak2/python/html-table-parser/scrape1.py', wdir='/Users/bhav 1/Documents/keepbak2/python/html-table-parser')

  File "/anaconda3/lib/python3.5/site-packages/spyder_kernels/customize/spydercustomize.py", line 668, in runfile
    execfile(filename, namespace)

  File "/anaconda3/lib/python3.5/site-packages/spyder_kernels/customize/spydercustomize.py", line 108, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)

  File "/Users/bhav 1/Documents/keepbak2/python/html-table-parser/scrape1.py", line 73, in <module>

  File "/anaconda3/lib/python3.5/site-packages/pandas/io/excel.py", line 1732, in save
    return self.book.close()

  File "/anaconda3/lib/python3.5/site-packages/xlsxwriter/workbook.py", line 306, in close

  File "/anaconda3/lib/python3.5/site-packages/xlsxwriter/workbook.py", line 635, in _store_workbook

  File "/anaconda3/lib/python3.5/zipfile.py", line 1009, in __init__
    self.fp = io.open(file, filemode)

FileNotFoundError: [Errno 2] No such file or directory: ''
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My bet is on the space in the string "bhav\ 1" to be a problem.
Try duplicating backslash llike this: "bhav\\ 1" in the XLS string path.
python is not a shell. So normally you do not have to (must not) escape spaces.

you should use '/Users/bhav 1/Documents/keepbak2/python/countrydiedtest.xlsx'

but to quickly test what's wrong I suggest to run following code snippet:

import os

fname = '/Users/bhav 1/Documents/keepbak2/python/countrydiedtest.xlsx'
while True:
    parent_dir = os.path.dirname(fname)
    if fname == parent_dir:
    if os.path.isdir(parent_dir):
        print("Directory %s exists" % parent_dir)
        print("Directory %s does not exist" % parent_dir)
    fname = parent_dir

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changing the file path resolved my issue.

#xlfile = 'countrydiedtest.xlsx'
xlfile = "/Users/bhav 1/Documents/dump/countrydiedtest.xlsx"

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This is what I suggested in my answer:

python is not a shell. So normally you do not have to (must not) escape spaces.

you should use '/Users/bhav 1/Documents/keepbak2/python/countrydiedtest.xlsx'

So I'm a little surprised, that you accept your and not my solution


Agreed. It was my error. Thanks.
no problem. Happy coding

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