def words(word,number):
if number<len(word):
result={}
for key,value in word.items():
common_num=sorted(set(word.values()), reverse=True)[:number]
if value in common_num:
result.update({key:value})
word.clear()
word.update(result)
new_word_count={}
common_word=[]
common=[]
for key, value in word.items():
if value in common_word:
common.append(value)
common_word.append(value)
new_word_count=dict(word)
for key,value in new_word_count.items():
if value in common:
del word[key]
Example:
>>> word={'a': 2, 'b': 2, 'c' : 3, 'd: 3, 'e': 4, 'f' : 4, 'g' : 5}
>>> words(word,3)
My output: {'g': 5}
Expected output:{'g': 5, 'e': 4, 'f': 4}
Any idea why im getting this output
Well, without any special imports, there are easier ways to accomplish what you're trying to do. You've got a whole lot of rigmarole involved in tracking and storing the values being kept, then deleting, then re-adding, when you could simplify a lot; even with explanatory comments, this is substantially shorter:
def common_words(word_count, number):
# Early out when no filtering needed
if number >= len(word_count):
return
# Get the top number+1 keys based on their values
top = sorted(word_count, key=word_count.get, reverse=True)[:number+1]
# We kept one more than we needed to figure out what the tie would be
tievalue = word_count[top.pop()]
# If there is a tie, we keep popping until the tied values are gone
while top and tievalue == word_count[top[-1]]:
top.pop()
# top is now the keys we should retain, easy to compute keys to delete
todelete = word_count.keys() - top
for key in todelete:
del word_count[key]
There are slightly better ways to do this that avoid repeated lookups in word_count
(sorting items
, not keys
, etc.), but this is easier to understand IMO, and the extra lookups in word_count
are bounded and linear, so it's not a big deal.
Although in the comments the author mentions avoiding Counter()
, for those interested in seeing how to apply it, here is a short solution as suggested by @ShadowRanger:
import collections as ct
word={'a': 2, 'b': 2, 'c' : 3, 'd': 3, 'e': 4, 'f' : 4, 'g' : 5}
words = ct.Counter(word)
words.most_common(3)
# [('g', 5), ('f', 4), ('e', 4)]
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