day6

day6

1.判断下标为是否为偶数的方式来实现，如果为耦合下标就把值累加

方法1

a = [1,2,1,2,3,3,3,3]

result = 0

for i in a[::2]:

result += i

方法2：

for i in range(len(a)):

if i%2 == 0:

result += a[i]

 

2.倒叙取出每个单词的第一个字母

>>> s = "i am a boy"

>>> s_list = s.split()

>>> s_list

['i', 'am', 'a', 'boy']

>>> for i in s_list:

... print(i[0])

 

3.找出 s=”aabbccddxxxxffff”中，出现次数最多的字母

result = {}

for i in s:

if i in result:

result[i] += 1

else:

result[i] = 1

 

4.自定义count函数

1.只统计单个字符出现的次数

def count(s,letter):

num = 0

if not isinstace(s,str) and not isinstace(letter,str):

return 0

else:

for i in s:

if i == letter:

num += 1

return num

2.统计多个字符出现的次数

def counts(s,letter):

num = 0

letter_len = len(letter)

if not isinstance(s,str) and not isinstace(letter,str):

rerurn 0

elif letter not in s:

return 0

else:

for i in range(len(s)):

if s[i:i+letter_len] == letter:

num += 1

rerutn num

5自定义divmod

def divmod(a,b):

c = a//b

d = a%b

return c,d