问题
if i just read my sum_digits
function here, it makes sense in my head but it seems to be producing wrong results. Any tip?
def is_a_digit(s):
''' (str) -> bool
Precondition: len(s) == 1
Return True iff s is a string containing a single digit character (between
'0' and '9' inclusive).
>>> is_a_digit('7')
True
>>> is_a_digit('b')
False
'''
return '0' <= s and s <= '9'
def sum_digits(digit):
b = 0
for a in digit:
if is_a_digit(a) == True:
b = int(a)
b += 1
return b
For the function sum_digits
, if i input sum_digits('hihello153john')
, it should produce 9
回答1:
Notice that you can easily solve this problem using built-in functions. This is a more idiomatic and efficient solution:
def sum_digits(digit):
return sum(int(x) for x in digit if x.isdigit())
sum_digits('hihello153john')
=> 9
In particular, be aware that the is_a_digit()
method already exists for string types, it's called isdigit().
And the whole loop in the sum_digits()
function can be expressed more concisely using a generator expression as a parameter for the sum()
built-in function, as shown above.
回答2:
You're resetting the value of b
on each iteration, if a
is a digit.
Perhaps you want:
b += int(a)
Instead of:
b = int(a)
b += 1
回答3:
Another way of using built in functions, is using the reduce function:
>>> numeric = lambda x: int(x) if x.isdigit() else 0
>>> reduce(lambda x, y: x + numeric(y), 'hihello153john', 0)
9
回答4:
One liner
sum_digits = lambda x: sum(int(y) for y in x if y.isdigit())
回答5:
I would like to propose a different solution using regx that covers two scenarios:
1.
Input = 'abcd45def05'
Output = 45 + 05 = 50
import re
print(sum(int(x) for x in re.findall(r'[0-9]+', my_str)))
Notice the '+' for one or more occurrences
2.
Input = 'abcd45def05'
Output = 4 + 5 + 0 + 5 = 14
import re
print(sum(int(x) for x in re.findall(r'[0-9]', my_str)))
回答6:
Another way of doing it:
def digit_sum(n):
new_n = str(n)
sum = 0
for i in new_n:
sum += int(i)
return sum
回答7:
An equivalent for your code, using list comprehensions:
def sum_digits(your_string):
return sum(int(x) for x in your_string if '0' <= x <= '9')
It will run faster then a "for" version, and saves a lot of code.
回答8:
Just a variation to @oscar's answer, if we need the sum to be single digit,
def sum_digits(digit):
s = sum(int(x) for x in str(digit) if x.isdigit())
if len(str(s)) > 1:
return sum_digits(s)
else:
return s
来源:https://stackoverflow.com/questions/14550034/sum-of-digits-in-a-string