态
以下是我添加整数的奇数时的尝试：
def sumOdd(n):
for i in range(n):
if n % 2 == 0: # if n is odd
n -= 1
print(sum(range(1, n, 2)) + n) # The range(1,n,2) starts at 1 and counts by twos until it reaches n
sumOdd(123) # 4
有什么提示吗？
怎么样：
代码：
def sum_odd_digits(number):
return sum(int(d) for d in str(number) if d in '13579')
测试代码：
print(sum_odd_digits(123))
print(sum_odd_digits(133))
结果：
4
7
def sumOdd(n):
n = str(n)
sumn = 0
for i in n:
i = int(i)
if i % 2 == 1: # if n is odd
sumn+=i
return sumn
print(sumOdd(132495)) # 4388797504
def sumOdd_(n):
n = abs(n)
sumn = 0
while n>0:
digit = n%10
n = n//10
if digit %2 ==1:
sumn+=digit
return sumn
myn = 132495
assert sumOdd_(myn)==sumOdd(myn)
否则，您可以使用Pythonic在Python中使用divmod的方法。并请注意，通常DIV和模式比投射到Str。
def sumOdd_2(n):
sumn=0
while n:
# "pop" the rightmost digit
n, digit = divmod(n, 10)
if digit %2 ==1:
sumn+=digit
return sumn
您也可以尝试以下方法：
数据的预处理：
data=123456789
real_data=list(map(int,str(data)))
在处理的数据上操作：
print(sum(filter(lambda x:x%2,real_data)))
或
print(functools.reduce(lambda x,y:x+y,(filter(lambda x:x%2,real_data))))
输出：
25
<pre>
def check_odd(a):
if a % 2 == 1:
return True
else:
return False
def extract_last_digit(a):
return a % 10
def remove_last_digit(a):
return a // 10
x = input('Type an integer: ')
n = int(x)
if n < 0: # the integer may be negative (alternatively use n = abs(int(x)) in previous line)
n = -1*n
sum_odd_n = 0
while n != 0:
if check_odd(n) == True:
sum_odd_n += extract_last_digit(n)
n = remove_last_digit(n)
print('The sum of the odd digits in number ', x, ' is ', str(sum_odd_n))
</pre>