Question

问题：可变句子存储一个字符串。写代码到确定句子中有多少个单词以相同的开头和结尾字母，包括一个字母的单词。将结果存储在变量中 same_letter_count。

我已经通过几种不同的方式进行了调整，但是我仍然无法弄清楚。感谢您的任何帮助和解释，以便我下次知道如何处理。

sentence = "students flock to the arb for a variety of outdoor activities 
such as jogging and picnicking"

same_letter_count = 0
sentence_split = sentence.split(' ')
sent_length = len(sentence_split)
#print(sent_length)
# Write your code here.
for d in sentence_split:
    #print(d[0])
    if d[0] == d:
        same_letter_count = same_letter_count + 1
    elif d[-1] == d:
        same_letter_count = same_letter_count + 1
print(same_letter_count)

我得到的答案是1，正确的答案是2。

Answer 1

您可以利用Python的布尔值可以被视为零和一的事实，并且只需将测试word[0] == word[-1]的所有布尔值相加即可。表达式：

[w[0] == w[-1] for w in sentence.split()]

评估为[True, False, False...]之类的列表。取其中的sum与计算True值的数量相同，是在Python中执行此类操作的一种非常典型的方法。

sentence = "students flock to the arb for a variety of outdoor activities such as jogging and picnicking"
same_letter_count = sum(w[0] == w[-1] for w in sentence.split())
# 2

Answer 2

if d[0] == d:
    same_letter_count = same_letter_count + 1
elif d[-1] == d:
    same_letter_count = same_letter_count + 1

这将检查第一个字母是否等于整个单词，或者最后一个字母是否等于整个单词。因此，您仅在计算“ a”。而是尝试

if d[0] == d[-1]:
    same_letter_count = same_letter_count + 1

Answer 3

pandas的快速解决方案：

s = pd.Series(sentence.split(' '))
(s.str[0] == s.str[-1]).sum()

给出答案2。

您甚至可以得到这些单词：

s[s.str[0] == s.str[-1]]

给予：

0    students
6           a
dtype: object

Answer 4

sentence = "students flock to the arb for a variety of outdoor activities such as jogging and picnicking"
same_letter_count = 0

sentence_split = sentence.split(' ')
sentense_length = len(sentence_split)

for d in sentence_split:
    if d[0] == d[-1]:
        same_letter_count += 1
print(same_letter_count)

计算多少个单词具有相同的开始和结束字母/ Python

4 个答案: