Python pandas 检查单元格中列表的最后一个元素是否包含特定字符串

Question

my dataframe df:

index                        url
1           [{'url': 'http://bhandarkarscollegekdp.org/'}]
2             [{'url': 'http://cateringinyourhome.com/'}]
3                                                     NaN
4                  [{'url': 'http://muddyjunction.com/'}]
5                       [{'url': 'http://ecskouhou.jp/'}]
6                     [{'url': 'http://andersrice.com/'}]
7       [{'url': 'http://durager.cz/'}, {'url': 'http:andersrice.com'}]
8            [{'url': 'http://milenijum-osiguranje.rs/'}]
9       [{'url': 'http://form-kind.org/'}, {'url': 'https://osiguranje'},{'url': 'http://beseka.com.tr'}]

I would like to select the rows if the last item in the list of the row of url column contains 'https', while skipping missing values.

My current script

df[df['url'].str[-1].str.contains('https',na=False)]

returns False values for all the rows while some of them actually contains https.

Can anybody help with it?

Answer 1

I think you can first replace NaN to empty url and then use apply :

df = pd.DataFrame({'url':[[{'url': 'http://bhandarkarscollegekdp.org/'}],
                          np.nan,
                         [{'url': 'http://cateringinyourhome.com/'}],  
                         [{'url': 'http://durager.cz/'}, {'url': 'https:andersrice.com'}]]},
                  index=[1,2,3,4])

print (df)
                                                 url
1     [{'url': 'http://bhandarkarscollegekdp.org/'}]
2                                                NaN
3        [{'url': 'http://cateringinyourhome.com/'}]
4  [{'url': 'http://durager.cz/'}, {'url': 'https...

---

df.loc[df.url.isnull(), 'url'] = [[{'url':''}]]
print (df)
                                                 url
1     [{'url': 'http://bhandarkarscollegekdp.org/'}]
2                                      [{'url': ''}]
3        [{'url': 'http://cateringinyourhome.com/'}]
4  [{'url': 'http://durager.cz/'}, {'url': 'https...

print (df.url.apply(lambda x: 'https' in x[-1]['url']))
1    False
2    False
3    False
4     True
Name: url, dtype: bool

First solution:

df.loc[df.url.notnull(), 'a'] = 
df.loc[df.url.notnull(), 'url'].apply(lambda x: 'https' in x[-1]['url'])

df.a.fillna(False, inplace=True)
print (df)
                                                 url      a
1     [{'url': 'http://bhandarkarscollegekdp.org/'}]  False
2                                                NaN  False
3        [{'url': 'http://cateringinyourhome.com/'}]  False
4  [{'url': 'http://durager.cz/'}, {'url': 'https...   True

Answer 2

not sure url is str or other types

you can do like this:

"https" in str(df.url[len(df)-1])

or

str(df.ix[len(df)-1].url).__contains__("https")