[英]How to add numbers in a string
我是 Python 新手,我有一个作业来验证信用卡号。 我完成了前两个条件,但我坚持 #3 和 #4 条件。 任何帮助表示赞赏
状况:
def verify(number) : # do not change this line!
# write your code here so that it verifies the card number
#condtion 1
if number[0] != '4':
return "violates rule #1"
#condition 2
if int(number[3]) != (int(number[5]) + 1) :
return "violates rule #2"
#condition 3
for i in number:
if i >= '0' and i !='-':
# be sure to indent your code!
return True # modify this line as needed
input = "4037-6000-0000" # change this as you test your function
output = verify(input) # invoke the method using a test input
print(output) # prints the output of the function
# do not remove this line!
预期产出:
● "5000-0000-0000": violates rule #1
● "4000-0000-0000": passes rule #1, violates rule #2
● "4007-6000-0000": passes rules #1-2, violates rule #3
● "4037-6000-0000": passes rules #1-3, violates rule #4
● “4094-3460-2754”: passes all rules
对于规则 3,您需要对所有数字求和并检查除以 4 的余数是否为零:
#condition 3
s = 0
for i in number:
if i != '-':
s += int(i)
if s % 4 != 0:
return "violates rule #3"
对于规则 4,您可以获得子字符串的 int 总和:
if (int(number[0:2]) + int(number[7:8])) != 100:
return "violates rule #4"
完整代码:
def verify(number) : # do not change this line!
# write your code here so that it verifies the card number
#condtion 1
if number[0] != '4':
return "violates rule #1"
#condition 2
if int(number[3]) != (int(number[5]) + 1) :
return "violates rule #2"
#condition 3
s = 0
for i in number:
if i != '-':
s += int(i)
if s % 4 != 0:
return "violates rule #3"
if (int(number[0:2]) + int(number[7:9])) != 100:
return "violates rule #4"
# be sure to indent your code!
return True # modify this line as needed
input = "4037-6000-0000" # change this as you test your function
output = verify(input) # invoke the method using a test input
print(output) # prints the output of the function
# do not remove this line!
对于条件 #3
Python 中的字符串是可迭代的,这意味着您可以在for
循环中传递它们。 循环中的每个元素都是字符串中的一个字符。 所以如果你做了
for char in "4094-3460-2754":
print(char)
你会得到:
4
0
9
4
-
3
4
etc.
使用它,您可以计算输入中每个数字的总和并检查它是否可以被 4 整除。需要注意的两件事,您需要首先将字符转换为整数(使用int
)。 你不能做
"4" + "0"
但是你可以做
int("4") + int("0")
您还需要使用if
从总和中排除“-”。
其次,我们使用模 ( %
) 检查两个数字在 Python 中是否可整除。 结果是余数,如果余数为 0,则第一个参数可被第二个参数整除。
16 % 4 == 0 # divisible by 4
21 % 4 == 1 # not divisible by 4
对于条件 #4
除了可迭代之外,Python 中的字符串还可以通过索引访问(从 0 开始)
"4094-3460-2754"[0] == "4"
"4094-3460-2754"[1] == "0"
"4094-3460-2754"[2] == "9"
"4094-3460-2754"[0:1] == "40"
"4094-3460-2754"[1:2] == "09"
因此,您可以访问多个字符并将它们视为整数:
int("4094-3460-2754"[0:1]) == 40
现在您可以将它们加在一起,看看它们是否等于 100。
所有数字之和必须能被 4 整除。
您可以使用条件语句进行检查,例如:
if sum_of_nums % 4 != 0:
print("Violates rule 3!")
这会检查除以 4 时是否有余数,如果没有余数,它将被平均除,并且表达式将等于 0。 如果它不均分,它将不等于 0!
如果将前两位数视为两位数,将第七位和第八位视为两位数,则它们的总和必须为 100。
在这里,您可以像引用列表一样引用字符串的字符。 如果输入始终保持一致,您可以对引用进行硬编码,将它们更改为整数,然后将它们加在一起并使用条件语句检查它们
first_num = input[0]+input[1]
first_num = int(first_num) #now an int
second_num = input[6]+input[7]
second_num = int(second_num)
if first_num + second_num != 100:
print("Violates rule 4!")
我宁愿去掉破折号-
从数量第一,这样它可以很容易地工作着。 您也可以在不删除它的情况下进行操作。
# split it into parts separated by dashes
# consider 4094-3460-2754
no_dashes = number.split('-')
print(no_dashes) # ['4094', '3460', '2754']
# combine the numbers without dashes
no_dashes = ''.join(no_dashes)
print(no_dashes) # 409434602754
# convert it into a list of integers so that it is more easier to work with
number = [int(x) for x in no_dashes]
print(number) # [4, 0, 9, 4, 3, 4, 6, 0, 2, 7, 5, 4]
现在,正如您所提到的,第一个条件很简单,您可以简单地检查第一个数字是否为 4。
# 1st condition
if number[0] != 4:
return 'Violates #1'
第二个条件也很简单:
# 2nd condition
# 4th digit is a[3] and 5th digit is a[4]
if number[3] != number[4] + 1:
return 'Viloates #2'
对于第三个条件,您只需找到数字中每个数字的总和。 由于我们已经将数字转换为整数数组,因此使用sum()
函数也很容易:
# 3rd condition
# Find the sum
num_sum = sum(number)
print(num_sum) # 48
# now check if the sum is divisible by 4
if num_sum % 4 != 0:
return 'Violates #3'
现在,对于第四个条件,您需要将第 1 位和第 2 位数字视为两位数,并与第 7 位和第 8 位数字相同。 您可以将其转换为两位数,如下所示:
# 1st digit is number[0]
# 2nd digit is number[1]
# 7th digit is number[6]
# 8ty digit is number [7]
# convert 1st two digits into a two-digit number
x = number[0] * 10 + number[1]
# convert 7th and 8th digits into a two-digit number
y = number[6] * 10 + number[7]
现在您可以检查它们的总和是否为 100:
if x + y != 100:
return 'Violates #4'
因此,组合程序变为(合并了一些步骤):
def verify(number):
number = [int(x) for x in ''.join(number.split('-'))]
if number[0] != 4:
return 'Violates #1'
if number[3] != number[4] + 1:
return 'Viloates #2'
if sum(number) % 4 != 0:
return 'Violates #3'
if (number[0] * 10 + number[1] + number[6] * 10 + number[7]) != 100:
return 'Violates #4'
return True
但是上面的程序只会给出失败的第一个条件。 如果需要,您可以进一步修改它,如下所示:
def verify(number):
failed = []
number = [int(x) for x in ''.join(number.split('-'))]
if number[0] != 4:
failed += [1]
if number[3] != number[4] + 1:
failed += [2]
if sum(number) % 4 != 0:
failed += [3]
if (number[0] * 10 + number[1] + number[6] * 10 + number[7]) != 100:
failed += [4]
res = 'Violates ' + (', '.join[str(x) for x in failed])
return res if len(failed) != 0 else 'Passed'
print(verify('4094-3460-2754')) # Passed
print(verify('4037-6000-0000')) # Violates 4
您也可以再次修改它以显示通过的条件。 我把它留给你!
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