how to make a dataframe by changing condition on url?
Question:
I have a dataframe, which contains many urls, is there a way i can make whenever it finds a specific url "https://abs.twimg.com/sticky/default_profile_images/default_profile_normal.png" it gives a value False, else true?
My dataframe:
Expected:
surname
value
First
True
Second
False
Third
True
Fourth
False
Answers:
Can you try the following:
url_pattern = "https://abs.twimg.com/sticky/default_profile_images/default_profile_normal.png"
df['value'] = df['image_url'].apply(lambda x: True if x == url_pattern else False)
You can use contains which creates a boolean mask. By assigning it to df['value'] you assign it to a new column, which gives you the desired result. This is even easier than creating a lambda function for it.
df['value'] = df.image_url.str.contains("https://abs.twimg.com/sticky/default_profile_images/default_profile_normal.png")
I have a dataframe, which contains many urls, is there a way i can make whenever it finds a specific url "https://abs.twimg.com/sticky/default_profile_images/default_profile_normal.png" it gives a value False, else true?
My dataframe:
Expected:
| surname | value |
|---|---|
| First | True |
| Second | False |
| Third | True |
| Fourth | False |
Can you try the following:
url_pattern = "https://abs.twimg.com/sticky/default_profile_images/default_profile_normal.png"
df['value'] = df['image_url'].apply(lambda x: True if x == url_pattern else False)
You can use contains which creates a boolean mask. By assigning it to df['value'] you assign it to a new column, which gives you the desired result. This is even easier than creating a lambda function for it.
df['value'] = df.image_url.str.contains("https://abs.twimg.com/sticky/default_profile_images/default_profile_normal.png")