Within the for loop, use an if statement to check whether the similarity score in each match is greater than or equal to

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小小草
小小草 LV1 2023年9月23日 23:40 发表
Here's the code to check whether the similarity score in each match is greater than or equal to 80, and if it is, use `.loc` to select rows where `cuisine_type` in `restaurants` is equal to the current match (which is the first element of `match`), and reassign them to be 'italian': ```python # Create a list of matches, comparing 'italian' with the cuisine_type column matches = process.extract('italian', restaurants['cuisine_type'], limit=len(restaurants.cuisine_type)) # Iterate through the list of matches to italian for match in matches: # Check whether the similarity score is greater than or equal to 80 if match[1] >= 80: # Select all rows where the cuisine_type is spelled this way, and set them to the correct cuisine restaurants.loc[restaurants['cuisine_type'] == match[0], 'cuisine_type'] = 'italian' ``` In this code, we use the `process.extract()` function from the `fuzzywuzzy` library to create a list of matches between 'italian' and the values in the `cuisine_type` column of the `restaurants` DataFrame. We pass 'italian' as the first argument, `restaurants['cuisine_type']` as the second argument, and `limit=len(restaurants.cuisine_type)` to ensure that we get a match score for every value in the `cuisine_type` column. We then iterate through the list of matches using a `for` loop. For each match, we check whether the similarity score (which is the second element of the `match` tuple) is greater than or equal to 80. If it is, we use `.loc` to select all rows where the `cuisine_type` is spelled this way (which is the first element of the `match` tuple), and set them to be 'italian'. We do this by passing two arguments to `.loc`: a boolean condition that selects the relevant rows (`restaurants['cuisine_type'] == match[0]`), and the name of the column we want to modify (`'cuisine_type'`). We then use the assignment operator (`=`) to set the selected rows to 'italian'.
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