Python KeyError字典

Question

• I have a list, called tweets_data
• Each element of the list is a dictionary
• Keys of the dictionary is 'text'
• But the raw data has some missing 'text'

That is why, I want to remove the dictionaries having missing text. This is how my code looks like:

for i in range(len(tweets_data)):
    try:
        print tweets_data[i]['text']
    except KeyError:
        tweets_data.remove(tweets_data[i])
        i += 1

And I am receiving such an error:

IndexError: list index out of range

My question: Is it possible to just remove the missing data from my list with a more clever way so that I won't get such an error? Thanks ahead!

Answer 1

You can't remove items from a list while you're iterating over it without confusing the indexes. Each time you remove, the list gets shorter - but you're still counting up to the length of the original list and expecting to find elements there.

Try this instead:

ok_tweets = [x for x in tweets_data if 'text' in x]

Answer 2

可能适合您采用其他方法

new_tweet_data = [tweet for tweet in tweet_data if 'text' in tweet]

Answer 3

I guess this one works...

cleandata=[]
for i in range(len(tweets_data)):
    try:
        print tweets_data[i]['text']
        cleandata.append(tweets_data[i]['text'])
    except KeyError:
        i += 1

Answer 4

If your data is of reasonable size I'd recommend a filtered list comprehension, as previously suggested by others

 
 
 
 
  
  
  filtered = [tweet for tweet in tweets_data if 'text' in tweet]
 
 
  

OTOH, if your list is LARGE and the defective items, those you want to remove, are just a few, it is possible that an approach based on .remove() may be faster, avoiding the intermediate step of creating a LARGE new list

 
 
 
 
  
  
  delenda = [defective for defective in tweet_data if 'text' not in defective] for tweet in delenda: tweeets_data.remove(tweet)
 
 
  

Beware that each .remove() has to scan the whole list, so this approach could be competitive only for a very small ratio of items to remove

If you need to deliver a product based on this question I heartily recommend timing the different approaches with samples of your data

Having read https://wiki.python.org/moin/TimeComplexity , in species

Internally, a list is represented as an array; the largest costs come from growing beyond the current allocation size (because everything must move), or from inserting or deleting somewhere near the beginning (because everything after that must move).

I have striked out my previous answer, suggesting the use of .remove() to avoid copying a possibly LARGE list, because it turns out that every .remove() is, in effect, COPYING a possibly large part of the list.

The right thing to do is indeed a list comprehension.