Parsing URLs in Pandas DataFrame
I don't know if it's still relevant for you, but I can see some optimization that you could do in your code.
As I see, you are passing all dataframe objects and then parsing specific columns in that dataframe:
def parse_url(df):
    for index, row in df.iterrows():
        parsed = urlparse(str(row['Destination URL'])).query #&lt;==
        parsed = parse_qs(parsed)
        for k, v in parsed.iteritems():
            df.loc[index, k.strip()] = v[0].strip().lower()
    return df

It would be faster if you pass only the column that you need to parse.
E.g.
def parse_url(df):
    for index, row in df.iterrows():
        parsed = urlparse(str(row)).query # &lt;- Notice the change here
        parsed = parse_qs(parsed)
        for k, v in parsed.items(): #use items() in Python3 and iteritems() in Python2
            df.loc[index, k.strip()] = v[0].strip().lower()
    return df


parse_url(df['columnName'])

Then first your function would have less work to do and performance would increase - at least slightly.