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[英]Counting how many times a word appears in a text using python and Counters
[英]Regular expressions - counting how many times a word appears in a text
我試圖設置的是一個函數,它給定某個文本將打印出單詞['color', 'Colour', 'Color','Colour']
出現的次數。 所以我得到以下結果:
assert colorcount("Color Purple") == 1
assert colorcount("Your colour is better than my colour") == 2
assert colorcount("color Color colour Colour") == 4
我擁有的是
import re
def colorcount(text):
all_matches = re.findall('color', 'Colour', 'Color'. 'Colour', text)
return len(all_matches)
print(colorcount(text)
它不起作用,那么我如何編寫代碼使其按照我的意願工作?
如果你想使用正則表達式,你可以這樣做:
import re
def colorcount(text):
r = re.compile(r'\bcolour\b | \bcolor\b', flags = re.I | re.X)
count = len(r.findall(text))
print(count)
return count
# These asserts work as expected without raising an AssertionError.
assert colorcount("Color Purple") == 1
assert colorcount("Your colour is better than my colour") == 2
assert colorcount("color Color colour Colour") == 4
哪些輸出:
1
2
4
您可以簡單地將文本轉換為特定大小寫(即全部較低),然后使用字符串的count()
來循環每次出現的關鍵字:
def colorcount(text):
KEY_WORDS = [ 'color', 'colour' ]
counter = 0
sanitexed_text = text.lower()
for kw in KEY_WORDS:
counter += sanitexed_text.count(kw.lower())
return counter
text = 'color ds das Colour dsafasft e re Color'
# 3
print(colorcount(text))
# All following asserts pass
assert colorcount("Color Purple") == 1
assert colorcount("Your colour is better than my colour") == 2
assert colorcount("color Color colour Colour") == 4
嘗試這個
def colorcount(text):
return len(re.findall('[c|C]olou{0,1}r', text))
assert colorcount("Color Purple") == 1
assert colorcount("Your colour is better than my colour") == 2
assert colorcount("color Color colour Colour") == 4
使用以下帶有標志re.I
(不區分大小寫)和re.findll
正則表達式,然后返回返回列表的長度:
\bcolou?r\b
import re
def colorcount(text):
return len(re.findall(r'\bcolou?r\b', text, flags=re.I))
print(colorcount('color Color colour Colour'))
印刷:
4
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