如果列值以pandas数据框开头/匹配字符串，则替换列值

Question

I have a column in my dataframe prices_df as thumbnail_url .

    zipcode thumbnail_url
0   11201   https://a0.muscache.com/im/pictures/6d7cbbf7-c...
1   10019   0
2   10027   https://a0.muscache.com/im/pictures/6fae5362-9...
3   94117   https://a0.muscache.com/im/pictures/72208dad-9...
4   20009   0
5   94131   https://a0.muscache.com/im/pictures/82509143-4...

I need to replace all values where the row contains https:// or lets say contains .com with numeric value 1.

zipcode thumbnail_url
0   11201   1
1   10019   0
2   10027   1

Tried this

img_Uploaded = prices_df['thumbnail_url'].str.contains("http") == True
prices_df.replace(to_replace=prices_df[img_Uploaded],value=1,inplace=True)

My dataframe is of shape (74111, 2)

This line of code takes too much time and my system froze. Can someone suggest a better vectorized operation and explain it.

My issue is resolved but I am curious what was wrong with my code ? Apart from the fact that it did not optimized using vectorized operations? It should still run right? Or THAT is the reason why it froze and did not run whereas the codes suggested below ran in seconds

Answer 1

您可以使用apply（）函数来完成此操作：

prices_df.thumbnail_url = prices_df.thumbnail_url.apply(lambda url: 1 if 'http' in str(url) else url)

Answer 2

You can use lambda expression

prices[['thumbnail_url']] = prices[['thumbnail_url']].apply(lambda x: 1 if 'https://' in str(x) else 0)

They are a shorthand to create anonymous functions; the expression lambda parameters: expression yields a function object. The unnamed object behaves like a function object defined with

here`s doc