CODEWITHSUNDEEP
pythonpandasdataframepython-requests
DNburtonguster
DNburtonguster

Reputation: 367

Construct URL query from pandas dataframe

Taking a pandas dataframe of the form:

fname lname age
mark  trumb 34
samb  jones 26

I'd like to construct a url query using the requests module. Normally, I would construct the url like so:

payload = {'key1':'value1','key2':'value2'}
headers = {'content-type': 'application'}
base_url = 'https://website.com/'
req = requests.post(base_url,headers=headers,params=payload)

How would I best use this so that the dataframe columns become the keys, and the rows become the values?

The new output column (req) would look like this:

fname lname age req
mark  trumb 34  https://website.com?fname=mark&lname=trumb&age=34
samb  jones 26  https://website.com?fname=samb&lname=jones&age=26

Upvotes: 2

Views: 1140

Answers (1)

EdChum
EdChum

Reputation: 394041

You can build it like this:

In [46]:
df['req'] = 'https://website.com?' + 'fname=' + df['fname'] + '&lname=' + df['lname'] + '&age=' + df['age'].astype(str)
df

Out[46]:
  fname  lname  age                                                req
0  mark  trumb   34  https://website.com?fname=mark&lname=trumb&age=34
1  samb  jones   26  https://website.com?fname=samb&lname=jones&age=26

A dynamic method can be done by joining the column names with the values:

In [106]:
def func(x):
    tuple_list = list(zip('&' + x.index + '=', x.astype(str)))
    return ''.join([''.join(item) for item in tuple_list])

df['req'] = r'https://website.com?' + df.apply(lambda x: func(x), axis=1)
df['req']

Out[106]:
0    https://website.com?&name=mark&lname=trumb&age=34
1    https://website.com?&name=samb&lname=jones&age=26
Name: req, dtype: object

Upvotes: 1

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