I want to scrape data at the county level from https://apidocs.covidactnow.org
However I could only get a dataframe with one line for each county, and data for each date is stored within a dictionary in each row/county. I would like to access this data and store it in long format (= have one row per county-date).
import requests
import pandas as pd
import os
if __name__ == '__main__':
os.chdir('/home/username/Desktop/')
url = 'https://api.covidactnow.org/v2/counties.timeseries.json?apiKey=ENTER_YOUR_KEY'
response = requests.get(url).json()
data = pd.DataFrame(response)
This seems like a trivial question, but I've tried for hours. What would be the best way to achieve that ?
Do you mean something like that?
import requests
url = 'https://api.covidactnow.org/v2/states.timeseries.csv?apiKey=YOURAPIKEY'
response = requests.get(url)
csv_response = (response.text)
# Then you can transform STRING to CSV
Check this fo string to CSV --> python parsing string to csv format
Related
I have a json file in the form
{"total_rows":1000,"rows":[{data},{data},{data}]}
and I just want
[{data},{data},{data}]
I know pandas has desired output to dataframe like:
import pandas as pd
file_reading = json.loads(open(url).read())
df = pd.DataFrame.from_dict(file_reading['rows'])
print(df)
But I am hoping for a way to do this outputting to json array and its a big dataset so I dont want to loop
You opened a file without closing it. There's nothing fancy needed, the JSON just translate into a dictionary in Python:
with open(url) as fp:
file_reading = json.load(fp)
df = pd.DataFrame(file_reading["rows"])
I would like to crawl several urls, while using the requests library in python. I am scrutinizing the GET requests as well as the response headers. However, when crawling and getting the data from different urls I am facing the problem, that I don't know all 'key:values', which are coming in. Thus writing those data to a valid csv file is not really possible, in my point of view. Therefore I want to write the data into a json file.
The problem is similar to the following thread from 2014, but not the same:
Get a header with Python and convert in JSON (requests - urllib2 - json)
import requests, json
urls = ['http://www.example.com/', 'http://github.com']
with open('test.json', 'w') as f:
for url in urls:
r = requests.get(url)
rh = r.headers
f.write(json.dumps(dict(rh), sort_keys=True, separators=(',', ':'), indent=4))
I expect a json file, with the headers for each URL. I get a Json file with those data, but my IDE (PyCHarm) is showing an Error, which states out that
JSON standard allows only one top-level value. I have read the documentation:https://docs.python.org/3/library/json.html#repeated-names-within-an-object; but did not get it. Any hint would be appreciated.
EDIT: The only thing which is missing in the outcome is another comma. But where do I enter it and what command do I need for this?
You need to add it to an array and then finally do the json dump to a file. This will work.
urls = ['http://www.example.com/', 'http://github.com']
headers = []
for url in urls:
r = requests.get(url)
header_dict = dict(r.headers)
header_dict['source_url'] = url
headers.append(header_dict)
with open('test.json', 'w', encoding='utf-8') as f:
json.dump(headers, f, sort_keys=True, separators=(',', ':'), indent=4)
You still can write it to a csv:
import pandas as pd
df = pd.DataFrame(headers)
df.to_csv('test.csv')
I have fetched some .json data from API.
import urllib2
test=urllib2.urlopen('url')
print test
How can I save it as a table or data frame? I am using Spark 2.0.
This is how I succeeded importing .json data from web into df:
from pyspark.sql import SparkSession, functions as F
from urllib.request import urlopen
spark = SparkSession.builder.getOrCreate()
url = 'https://web.url'
jsonData = urlopen(url).read().decode('utf-8')
rdd = spark.sparkContext.parallelize([jsonData])
df = spark.read.json(rdd)
For this you can have some research and try using sqlContext. Here is Sample code:-
>>> df2 = sqlContext.jsonRDD(test)
>>> df2.first()
Moreover visit line and check for more things here,
https://spark.apache.org/docs/1.6.2/api/python/pyspark.sql.html
Adding to Rakesh Kumar answer, the way to do it in spark 2.0 is:
http://spark.apache.org/docs/2.1.0/sql-programming-guide.html#data-sources
As an example, the following creates a DataFrame based on the content of a JSON file:
# spark is an existing SparkSession
df = spark.read.json("examples/src/main/resources/people.json")
# Displays the content of the DataFrame to stdout
df.show()
Note that the file that is offered as a json file is not a typical JSON file. Each line must contain a separate, self-contained valid JSON object. For more information, please see JSON Lines text format, also called newline-delimited JSON. As a consequence, a regular multi-line JSON file will most often fail.
from pyspark import SparkFiles
from pyspark.sql import SparkSession
spark = SparkSession.builder.appName("Project").getOrCreate()
zip_url = "https://raw.githubusercontent.com/spark-examples/spark-scala-examples/master/src/main/resources/zipcodes.json"
spark.sparkContext.addFile(zip_url)
zip_df = spark.read.json("file://" +SparkFiles.get("zipcodes.json"))
#click on raw and then copy url
I am trying to develop an application to retrieve stock prices (in JSON) and then do some analysis on them. My problem is with getting the JSON response into a pandas DataFrame where I can work. Here is my code:
'''
References
http://stackoverflow.com/questions/6862770/python-3-let-json-object- accept-bytes-or-let-urlopen-output-strings
'''
import json
import pandas as pd
from urllib.request import urlopen
#set API call
url = "https://www.quandl.com/api/v3/datasets/WIKI/AAPL.json?start_date=2017-01-01&end_date=2017-01-31"
#make call and receive response
response = urlopen(url).read().decode('utf8')
dataresponse = json.loads(response)
#check incoming
#print(dataresponse)
df = pd.read_json(dataresponse)
print(df)
The application errors at df = pd.read_json... with error TypeError: Expected String or Unicode.
So I reckon this is the first hurdle.
The second is getting where I need to. The JSON response contains only two arrays I am interested in, column_names and data. How do I extract only these two and put into a pandas DataFrame?
To answer your first question, pd.read_json takes a JSON string directly, so you should be doing this:
pd.read_json(response)
But instead, considering how the data is structured, it's best to first convert the JSON string to a dictionary containing the data:
d = json.loads(response)
Then simply build the dataframe from d['dataset']['data'] and d['dataset']['column_names']:
pd.DataFrame(data=d['dataset']['data'], columns=d['dataset']['column_names'])
I want to manipulate the information at THIS url. I can successfully open it and read its contents. But what I really want to do is throw out all the stuff I don't want, and to manipulate the stuff I want to keep.
Is there a way to convert the string into a dict so I can iterate over it? Or do I just have to parse it as is (str type)?
from urllib.request import urlopen
url = 'http://www.quandl.com/api/v1/datasets/FRED/GDP.json'
response = urlopen(url)
print(response.read()) # returns string with info
When I printed response.read() I noticed that b was preprended to the string (e.g. b'{"a":1,..). The "b" stands for bytes and serves as a declaration for the type of the object you're handling. Since, I knew that a string could be converted to a dict by using json.loads('string'), I just had to convert the byte type to a string type. I did this by decoding the response to utf-8 decode('utf-8'). Once it was in a string type my problem was solved and I was easily able to iterate over the dict.
I don't know if this is the fastest or most 'pythonic' way of writing this but it works and theres always time later of optimization and improvement! Full code for my solution:
from urllib.request import urlopen
import json
# Get the dataset
url = 'http://www.quandl.com/api/v1/datasets/FRED/GDP.json'
response = urlopen(url)
# Convert bytes to string type and string type to dict
string = response.read().decode('utf-8')
json_obj = json.loads(string)
print(json_obj['source_name']) # prints the string with 'source_name' key
You can also use python's requests library instead.
import requests
url = 'http://www.quandl.com/api/v1/datasets/FRED/GDP.json'
response = requests.get(url)
dict = response.json()
Now you can manipulate the "dict" like a python dictionary.
json works with Unicode text in Python 3 (JSON format itself is defined only in terms of Unicode text) and therefore you need to decode bytes received in HTTP response. r.headers.get_content_charset('utf-8') gets your the character encoding:
#!/usr/bin/env python3
import io
import json
from urllib.request import urlopen
with urlopen('https://httpbin.org/get') as r, \
io.TextIOWrapper(r, encoding=r.headers.get_content_charset('utf-8')) as file:
result = json.load(file)
print(result['headers']['User-Agent'])
It is not necessary to use io.TextIOWrapper here:
#!/usr/bin/env python3
import json
from urllib.request import urlopen
with urlopen('https://httpbin.org/get') as r:
result = json.loads(r.read().decode(r.headers.get_content_charset('utf-8')))
print(result['headers']['User-Agent'])
TL&DR: When you typically get data from a server, it is sent in bytes. The rationale is that these bytes will need to be 'decoded' by the recipient, who should know how to use the data. You should decode the binary upon arrival to not get 'b' (bytes) but instead a string.
Use case:
import requests
def get_data_from_url(url):
response = requests.get(url_to_visit)
response_data_split_by_line = response.content.decode('utf-8').splitlines()
return response_data_split_by_line
In this example, I decode the content that I received into UTF-8. For my purposes, I then split it by line, so I can loop through each line with a for loop.
I guess things have changed in python 3.4. This worked for me:
print("resp:" + json.dumps(resp.json()))