I am trying to use the tablib library and create a Dataset from a .csv file. The following works:
import tablib
dataset = tablib.Dataset().load(open('data.csv').read())
However, in some cases, I'd like to load the .csv file from a URL.
Any ideas on how to do that?
You wrote
def get_ds(filename):
return tablib.Dataset().load(open(filename).read())
You want
import os.path
import requests
def get_ds(src):
if os.path.exists(src):
txt = open(src).read()
else:
req = requests.get(src)
req.raise_for_status()
txt = req.text
return tablib.Dataset().load(txt)
I am trying to create an API function, that takes in .csv file (uploaded) and opens it as pandas DataFrame. Like that:
from fastapi import FastAPI
from fastapi import UploadFile, Query, Form
import pandas as pd
app = FastAPI()
#app.post("/check")
def foo(file: UploadFile):
df = pd.read_csv(file.file)
return len(df)
Then, I am invoking my API:
import requests
url = 'http://127.0.0.1:8000/check'
file = {'file': open('data/ny_pollution_events.csv', 'rb')}
resp = requests.post(url=url, files=file)
print(resp.json())
But I got such error: FileNotFoundError: [Errno 2] No such file or directory: 'ny_pollution_events.csv'
As far as I understand from doc pandas is able to read .csv file from file-like object, which file.file is supposed to be. But it seems, that here in read_csv() method pandas obtains name (not a file object itself) and tries to find it locally.
Am I doing something wrong?
Can I somehow implement this logic?
To read the file in pandas, the file must be stored on your PC. Don't forget to import shutil. if you don't need the file to be stored on your PC, delete it using os.remove(filepath).
if not file.filename.lower().endswith(('.csv',".xlsx",".xls")):
return 404,"Please upload xlsx,csv or xls file."
if file.filename.lower().endswith(".csv"):
extension = ".csv"
elif file.filename.lower().endswith(".xlsx"):
extension = ".xlsx"
elif file.filename.lower().endswith(".xls"):
extension = ".xls"
# eventid = datetime.datetime.now().strftime('%Y%m-%d%H-%M%S-') + str(uuid4())
filepath = "location where you want to store file"+ extension
with open(filepath, "wb") as buffer:
shutil.copyfileobj(file.file, buffer)
try:
if filepath.endswith(".csv"):
df = pd.read_csv(filepath)
else:
df = pd.read_excel(filepath)
except:
return 401, "File is not proper"
I have managed to get my first python script to work which downloads a list of .ZIP files from a URL and then proceeds to extract the ZIP files and writes them to disk.
I am now at a loss to achieve the next step.
My primary goal is to download and extract the zip file and pass the contents (CSV data) via a TCP stream. I would prefer not to actually write any of the zip or extracted files to disk if I could get away with it.
Here is my current script which works but unfortunately has to write the files to disk.
import urllib, urllister
import zipfile
import urllib2
import os
import time
import pickle
# check for extraction directories existence
if not os.path.isdir('downloaded'):
os.makedirs('downloaded')
if not os.path.isdir('extracted'):
os.makedirs('extracted')
# open logfile for downloaded data and save to local variable
if os.path.isfile('downloaded.pickle'):
downloadedLog = pickle.load(open('downloaded.pickle'))
else:
downloadedLog = {'key':'value'}
# remove entries older than 5 days (to maintain speed)
# path of zip files
zipFileURL = "http://www.thewebserver.com/that/contains/a/directory/of/zip/files"
# retrieve list of URLs from the webservers
usock = urllib.urlopen(zipFileURL)
parser = urllister.URLLister()
parser.feed(usock.read())
usock.close()
parser.close()
# only parse urls
for url in parser.urls:
if "PUBLIC_P5MIN" in url:
# download the file
downloadURL = zipFileURL + url
outputFilename = "downloaded/" + url
# check if file already exists on disk
if url in downloadedLog or os.path.isfile(outputFilename):
print "Skipping " + downloadURL
continue
print "Downloading ",downloadURL
response = urllib2.urlopen(downloadURL)
zippedData = response.read()
# save data to disk
print "Saving to ",outputFilename
output = open(outputFilename,'wb')
output.write(zippedData)
output.close()
# extract the data
zfobj = zipfile.ZipFile(outputFilename)
for name in zfobj.namelist():
uncompressed = zfobj.read(name)
# save uncompressed data to disk
outputFilename = "extracted/" + name
print "Saving extracted file to ",outputFilename
output = open(outputFilename,'wb')
output.write(uncompressed)
output.close()
# send data via tcp stream
# file successfully downloaded and extracted store into local log and filesystem log
downloadedLog[url] = time.time();
pickle.dump(downloadedLog, open('downloaded.pickle', "wb" ))
Below is a code snippet I used to fetch zipped csv file, please have a look:
Python 2:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(StringIO(resp.read()))
for line in myzip.open(file).readlines():
print line
Python 3:
from io import BytesIO
from zipfile import ZipFile
from urllib.request import urlopen
# or: requests.get(url).content
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(BytesIO(resp.read()))
for line in myzip.open(file).readlines():
print(line.decode('utf-8'))
Here file is a string. To get the actual string that you want to pass, you can use zipfile.namelist(). For instance,
resp = urlopen('http://mlg.ucd.ie/files/datasets/bbc.zip')
myzip = ZipFile(BytesIO(resp.read()))
myzip.namelist()
# ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
My suggestion would be to use a StringIO object. They emulate files, but reside in memory. So you could do something like this:
# get_zip_data() gets a zip archive containing 'foo.txt', reading 'hey, foo'
import zipfile
from StringIO import StringIO
zipdata = StringIO()
zipdata.write(get_zip_data())
myzipfile = zipfile.ZipFile(zipdata)
foofile = myzipfile.open('foo.txt')
print foofile.read()
# output: "hey, foo"
Or more simply (apologies to Vishal):
myzipfile = zipfile.ZipFile(StringIO(get_zip_data()))
for name in myzipfile.namelist():
[ ... ]
In Python 3 use BytesIO instead of StringIO:
import zipfile
from io import BytesIO
filebytes = BytesIO(get_zip_data())
myzipfile = zipfile.ZipFile(filebytes)
for name in myzipfile.namelist():
[ ... ]
I'd like to offer an updated Python 3 version of Vishal's excellent answer, which was using Python 2, along with some explanation of the adaptations / changes, which may have been already mentioned.
from io import BytesIO
from zipfile import ZipFile
import urllib.request
url = urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/loc162txt.zip")
with ZipFile(BytesIO(url.read())) as my_zip_file:
for contained_file in my_zip_file.namelist():
# with open(("unzipped_and_read_" + contained_file + ".file"), "wb") as output:
for line in my_zip_file.open(contained_file).readlines():
print(line)
# output.write(line)
Necessary changes:
There's no StringIO module in Python 3 (it's been moved to io.StringIO). Instead, I use io.BytesIO]2, because we will be handling a bytestream -- Docs, also this thread.
urlopen:
"The legacy urllib.urlopen function from Python 2.6 and earlier has been discontinued; urllib.request.urlopen() corresponds to the old urllib2.urlopen.", Docs and this thread.
Note:
In Python 3, the printed output lines will look like so: b'some text'. This is expected, as they aren't strings - remember, we're reading a bytestream. Have a look at Dan04's excellent answer.
A few minor changes I made:
I use with ... as instead of zipfile = ... according to the Docs.
The script now uses .namelist() to cycle through all the files in the zip and print their contents.
I moved the creation of the ZipFile object into the with statement, although I'm not sure if that's better.
I added (and commented out) an option to write the bytestream to file (per file in the zip), in response to NumenorForLife's comment; it adds "unzipped_and_read_" to the beginning of the filename and a ".file" extension (I prefer not to use ".txt" for files with bytestrings). The indenting of the code will, of course, need to be adjusted if you want to use it.
Need to be careful here -- because we have a byte string, we use binary mode, so "wb"; I have a feeling that writing binary opens a can of worms anyway...
I am using an example file, the UN/LOCODE text archive:
What I didn't do:
NumenorForLife asked about saving the zip to disk. I'm not sure what he meant by it -- downloading the zip file? That's a different task; see Oleh Prypin's excellent answer.
Here's a way:
import urllib.request
import shutil
with urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/2015-2_UNLOCODE_SecretariatNotes.pdf") as response, open("downloaded_file.pdf", 'w') as out_file:
shutil.copyfileobj(response, out_file)
I'd like to add my Python3 answer for completeness:
from io import BytesIO
from zipfile import ZipFile
import requests
def get_zip(file_url):
url = requests.get(file_url)
zipfile = ZipFile(BytesIO(url.content))
files = [zipfile.open(file_name) for file_name in zipfile.namelist()]
return files.pop() if len(files) == 1 else files
write to a temporary file which resides in RAM
it turns out the tempfile module ( http://docs.python.org/library/tempfile.html ) has just the thing:
tempfile.SpooledTemporaryFile([max_size=0[,
mode='w+b'[, bufsize=-1[, suffix=''[,
prefix='tmp'[, dir=None]]]]]])
This
function operates exactly as
TemporaryFile() does, except that data
is spooled in memory until the file
size exceeds max_size, or until the
file’s fileno() method is called, at
which point the contents are written
to disk and operation proceeds as with
TemporaryFile().
The resulting file has one additional
method, rollover(), which causes the
file to roll over to an on-disk file
regardless of its size.
The returned object is a file-like
object whose _file attribute is either
a StringIO object or a true file
object, depending on whether
rollover() has been called. This
file-like object can be used in a with
statement, just like a normal file.
New in version 2.6.
or if you're lazy and you have a tmpfs-mounted /tmp on Linux, you can just make a file there, but you have to delete it yourself and deal with naming
Adding on to the other answers using requests:
# download from web
import requests
url = 'http://mlg.ucd.ie/files/datasets/bbc.zip'
content = requests.get(url)
# unzip the content
from io import BytesIO
from zipfile import ZipFile
f = ZipFile(BytesIO(content.content))
print(f.namelist())
# outputs ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
Use help(f) to get more functions details for e.g. extractall() which extracts the contents in zip file which later can be used with with open.
All of these answers appear too bulky and long. Use requests to shorten the code, e.g.:
import requests, zipfile, io
r = requests.get(zip_file_url)
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall("/path/to/directory")
Vishal's example, however great, confuses when it comes to the file name, and I do not see the merit of redefing 'zipfile'.
Here is my example that downloads a zip that contains some files, one of which is a csv file that I subsequently read into a pandas DataFrame:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
import pandas
url = urlopen("https://www.federalreserve.gov/apps/mdrm/pdf/MDRM.zip")
zf = ZipFile(StringIO(url.read()))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
(Note, I use Python 2.7.13)
This is the exact solution that worked for me. I just tweaked it a little bit for Python 3 version by removing StringIO and adding IO library
Python 3 Version
from io import BytesIO
from zipfile import ZipFile
import pandas
import requests
url = "https://www.nseindia.com/content/indices/mcwb_jun19.zip"
content = requests.get(url)
zf = ZipFile(BytesIO(content.content))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
It wasn't obvious in Vishal's answer what the file name was supposed to be in cases where there is no file on disk. I've modified his answer to work without modification for most needs.
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
def unzip_string(zipped_string):
unzipped_string = ''
zipfile = ZipFile(StringIO(zipped_string))
for name in zipfile.namelist():
unzipped_string += zipfile.open(name).read()
return unzipped_string
Use the zipfile module. To extract a file from a URL, you'll need to wrap the result of a urlopen call in a BytesIO object. This is because the result of a web request returned by urlopen doesn't support seeking:
from urllib.request import urlopen
from io import BytesIO
from zipfile import ZipFile
zip_url = 'http://example.com/my_file.zip'
with urlopen(zip_url) as f:
with BytesIO(f.read()) as b, ZipFile(b) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read())
If you already have the file downloaded locally, you don't need BytesIO, just open it in binary mode and pass to ZipFile directly:
from zipfile import ZipFile
zip_filename = 'my_file.zip'
with open(zip_filename, 'rb') as f:
with ZipFile(f) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read().decode('utf-8'))
Again, note that you have to open the file in binary ('rb') mode, not as text or you'll get a zipfile.BadZipFile: File is not a zip file error.
It's good practice to use all these things as context managers with the with statement, so that they'll be closed properly.
So an update, I found my compile issue was that I needed to change my notebook to a py file and choosing save as doesn't do that. So I had to run a different script turn my notebook to a py file. And part of my exe issue was I was using the fopen command that apparently isn't useable when compiled into a exe. So I redid the code to what is above. But now I get a write error when trying to run the script. I can not find anything on write functions with os is there somewhere else I should look?
Original code:
import requests
import json
import pandas as pd
import csv
from pathlib import Path
response = requests.get('url', headers={'CERT': 'cert'}, stream=True).json()
json2 = json.dumps(response)
f = open('data.json', 'r+')
f.write(json2)
f.close()
Path altered code:
import requests
import json
import pandas as pd
import csv
from pathlib import Path
response = requests.get('url', headers={'CERT': 'cert'}, stream=True).json()
json2 = json.dumps(response)
filename = 'data.json'
if '_MEIPASS2' in os.environ:
filename = os.path.join(os.environ['_MEIPASS2'], filename)
fd = open(filename, 'r+')
fd.write(json2)
fd.close()
The changes to the code allowed me to get past the fopen issue but created a write issue. Any ideas?
If you want to write to a file, you have to open it as writable.
fd = open(filename, 'wb')
Although I don't know why you're opening it in binary if you're writing text.
What I need is a command line tool to convert excel and ods spreadsheet files to csv which I can use on a web server (Ubuntu 16.04).
I already red this: https://pypi.python.org/pypi/unotools
which works fine for the given examples.
And this: http://www.linuxjournal.com/content/convert-spreadsheets-csv-files-python-and-pyuno-part-1v2
which should do the work I want it to do, but does not in my environment.
My problem I think is in the method Calc.store_to_url:
Line throwing exception
component.store_to_url(url,'FilterName','Text - txt - csv (StarCalc)')
I really would appreciate a hint.
Exception
unotools.unohelper.ErrorCodeIOException: SfxBaseModel::impl_store failed: 0x81a
Full source
import sys
from os.path import basename, join as pathjoin, splitext
from unotools import Socket, connect
from unotools.component.calc import Calc
from unotools.unohelper import convert_path_to_url
from unotools import parse_argument
def get_component(args, context):
_, ext = splitext(args.file_)
url = convert_path_to_url(args.file_)
component = Calc(context, url)
return component
def convert_csv(args, context):
component = get_component(args, context)
url = 'out/result.csv'
component.store_to_url(url,'FilterName','Text - txt - csv (StarCalc)')
component.close(True)
args = parse_argument(sys.argv[1:])
context = connect(Socket(args.host, args.port), option=args.option)
convert_csv(args, context)
The URL must be in file:// format.
url = convert_path_to_url('out/result.csv')
See the store_to_url example at https://pypi.python.org/pypi/unotools.
EDIT:
To use the absolute path, choose one of these; there is no need to combine them.
url = 'file:///home/me/out/result.csv'
url = convert_path_to_url('/home/me/out/result.csv')
To use the relative path, first verify that the working directory is '/home/me' by calling os.getcwd().