我目前正在學習Python，我需要編寫一個程序來確定一首詩中出現(xiàn)次數(shù)最多的單詞。使我困擾的問題是將一首詩的行解析成一個包含詩詞的單列表。當我解決它時，我將毫不費力地確定出現(xiàn)次數(shù)最多的單詞。我可以通過反復(fù)調(diào)用input（）來訪問這首詩的行，最后一行包含三個字符###。所以，我寫道：while True: y = input() if y == "###": break y = y.lower() y = y.split()并輸入：Here is a line like sparkling wineLine up now behind the cow###得到了結(jié)果：['here', 'is', 'a', 'line', 'like', 'sparkling', 'wine']['line', 'up', 'now', 'behind', 'the', 'cow']如果我嘗試調(diào)用y [0]，則會得到：hereline如何在同一變量內(nèi)連接兩個列表，或者如何將每一行分配給不同的變量？
2 回答
阿波羅的戰(zhàn)車
TA貢獻1862條經(jīng)驗 獲得超6個贊
words = []
while True:
y = input()
if y == "###":
break
words.extend(y.lower().split())
from collections import Counter
Counter(words).most_common(1)
整個代碼可以壓縮為以下一種格式：
Counter(y.lower() for x in iter(input, '###') for y in x.split()).most_common(1)
例如。
>>> sys.stdin = StringIO.StringIO("""Here is a line like sparkling wine
Line up now behind the cow
###""")
>>> Counter(y.lower() for x in iter(input, '###') for y in x.split()).most_common(1)
[('line', 2)]
由于每個請求，而沒有任何import小號
c = {} # stores counts
for line in iter(input, '###'):
for word in line.lower().split():
c[word] = c.get(word, 0) + 1 # gets count of word or 0 if doesn't exist
print(max(c, key=c.get)) # gets max key of c, based on val (count)
不必再次通過字典來查找最大單詞，請在進行操作時始終對其進行跟蹤：
c = {} # stores counts
max_word = None
for line in iter(input, '###'):
for word in line.lower().split():
c[word] = c.get(word, 0) + 1 # gets count of word or 0 if doesn't exist
if max_word is None:
max_word = word
else:
max_word = max(max_word, word, key=c.get) # max based on count
print(max_word)
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