Silent Erroer Handling In Python?

I got csv-file with numerous URLs. I read it into a pandas dataframe for convenience. I need to do some statistical work later - and pandas is just handy. It looks a little like th

Solution 1:

Try defining a function that does the "car" checking first and the use the .apply method of a pandas Series to get your 1, 0 or Wrong URL. The following should help:

import pandas as pd
import requests


data = [{"URLs" : "https://www.mercedes-benz.de", "electric" : 1},
        {"URLs" : "https://www.audi.de", "electric" : 0}, 
        {"URLs" : "https://ww.audo.e", "electric" : 0}, 
        {"URLs" : "NaN", "electric" : 0}]


defcontains_car(link):
    try:
        returnint('car'in requests.get(link).text)
    except:
        return"Wrong/Missing URL"


df = pd.DataFrame(data)

df['extra_column'] = df.URLs.apply(contains_car)


#                           URLs  electric extra_column# 0  https://www.mercedes-benz.de         1            1# 1           https://www.audi.de         0            1# 2             https://ww.audo.e         0    Wrong/Missing URL# 3                           NaN         0    Wrong/Missing URL

Edit:

You can search for more than just one keyword in the returned text from your HTTP request. Depending on the condition you set up, this can be done with either the builtin function any or the builtin function all. Using any means that finding any of the keywords should return 1, while using all means that all the keywords have to be matched in order to return 1. In the following example, I am using any with keywords such as 'car', 'automobile', 'vehicle':

import pandas as pd
import requests


data = [{"URLs" : "https://www.mercedes-benz.de", "electric" : 1},
        {"URLs" : "https://www.audi.de", "electric" : 0}, 
        {"URLs" : "https://ww.audo.e", "electric" : 0}, 
        {"URLs" : "NaN", "electric" : 0}]


defcontains_keywords(link, keywords):
    try:
        output = requests.get(link).text
        returnint(any(x in output for x in keywords))
    except:
        return"Wrong/Missing URL"


df = pd.DataFrame(data)
mykeywords = ('car', 'vehicle', 'automobile')
df['extra_column'] = df.URLs.apply(lambda l: contains_keywords(l, mykeywords))

Should yield:

#                            URLs  electric       extra_column# 0  https://www.mercedes-benz.de         1                  1# 1           https://www.audi.de         0                  1# 2             https://ww.audo.e         0  Wrong/Missing URL# 3                           NaN         0  Wrong/Missing URL

I hope this helps.

Solution 2:

I hope I do get you right, that 'NaN' is a "wrong/missing" URL. In this case you can just check for that. There are endless ways to indicate a missing URL. I'd prefere a missing value for car: Try this:

import pandas as pd
csv = [{"URLs" : "www.mercedes-benz.de", "electric" : 1}, {"URLs" : "www.audi.de", "electric" : 0}, {"URLs" : "ww.audo-car.e", "electric" : 0}, {"URLs" : "NaN", "electric" : 0}]
df = pd.DataFrame(csv)

print(df)

for i, row in df.iterrows():
    page_content = row['URLs']
    if page_content isNoneor page_content is"NaN":
        df.loc[i, 'car'] = Noneelif"car"in page_content:
        df.loc[i, 'car'] = Trueelse:
        df.loc[i, 'car'] = Falseprint(df.loc[i, 'car'])

print(df)

I edited some more things in your code, as they did not work. E.g this line with page_content = requests.get(row['URLs']) - requests is not defined. I guess you mean row.