CheckiO 是面向初学者和高级程序员的编码游戏,使用 Python 和 JavaScript 解决棘手的挑战和有趣的任务,从而提高你的编码技能,本博客主要记录自己用 Python 在闯关时的做题思路和实现代码,同时也学习学习其他大神写的代码。
CheckiO 官网:https://checkio.org/
我的 CheckiO 主页:https://py.checkio.org/user/TRHX/
CheckiO 题解系列专栏:https://itrhx.blog.csdn.net/category_9536424.html
CheckiO 所有题解源代码:https://github.com/TRHX/Python-CheckiO-Exercise
题目描述
【Popular Words】:给定两个参数,第一个为字符串(text
),第二个为字符串组成的列表(words
),要求计算该列表各个元素在第一个字符串中出现的次数,不区分大小写,如果一次都没出现,则为 0,返回结果为一个字典。
【链接】:https://py.checkio.org/mission/popular-words/
【输入】:一个字符串和要搜索的词组成的列表
【输出】:字典,其中搜索词是键,值是这些词在给定文本中出现的次数
【前提】:输入文本将由大小写英文字母和空格组成
【范例】:
popular_words('''
When I was One
I had just begun
When I was Two
I was nearly new
''', ['i', 'was', 'three', 'near']) == {
'i': 4,
'was': 3,
'three': 0,
'near': 0
}
解题思路
先将原字符串所有字母转换成小写字母,使用 split()
方法对其进行分割,分割后的每个单词作为元素,组成一个新列表,然后利用 count()
方法计算 words
中每个单词在 text
中出现的次数,将搜索词作为键,将其出现的次数作为值,存入到字典当中即可。
代码实现
def popular_words(text: str, words: list) -> dict:
text = text.lower().split()
dic = {}
for i in words:
dic[i] = text.count(i)
return dic
if __name__ == '__main__':
print("Example:")
print(popular_words('''
When I was One
I had just begun
When I was Two
I was nearly new
''', ['i', 'was', 'three', 'near']))
# These "asserts" are used for self-checking and not for an auto-testing
assert popular_words('''
When I was One
I had just begun
When I was Two
I was nearly new
''', ['i', 'was', 'three', 'near']) == {
'i': 4,
'was': 3,
'three': 0,
'near': 0
}
print("Coding complete? Click 'Check' to earn cool rewards!")
大神解答
大神解答 NO.1
def popular_words(text, words):
lower_count = text.lower().split().count
return {word: lower_count(word) for word in words}
大神解答 NO.2
def popular_words(text, words):
return dict(zip(words, map(text.lower().split().count, words)))
大神解答 NO.3
def popular_words(text: str, words):
res = {}
text = " "+text.lower().replace("n", " ")+" "
for i in words:
x, c = -1, 0
while True:
x = text.find(" "+i+" ", x+1)
if x != -1:
c += 1
else:
res.update({i: c})
break
return res
最后
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