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Python integer decimal

将整数拆分为数字以计算 ISBN 校验和

发布于 2024-07-24 04:30:51 字数 144 浏览 6 评论 0原文

我正在编写一个程序来计算 ISBN 号码的校验位。我必须将用户的输入（ISBN 的九位数字）读入整数变量，然后将最后一位数字乘以 2，倒数第二位数字乘以 3，依此类推。如何将整数“拆分”为其组成数字来做到这一点？由于这是一项基本的家庭作业练习，我不应该使用列表。

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混吃等死 2024-07-31 04:30:51

只需从中创建一个字符串即可。

myinteger = 212345
number_string = str(myinteger)

这就够了。现在您可以迭代它：

for ch in number_string:
    print ch # will print each digit in order

或者您可以对其进行切片：

print number_string[:2] # first two digits
print number_string[-3:] # last three digits
print number_string[3] # forth digit

或者更好的是，不要将用户的输入转换为整数（用户键入字符串）

isbn = raw_input()
for pos, ch in enumerate(reversed(isbn)):
    print "%d * %d is %d" % pos + 2, int(ch), int(ch) * (pos + 2)

有关更多信息，请阅读教程。

Just create a string out of it.

myinteger = 212345
number_string = str(myinteger)

That's enough. Now you can iterate over it:

for ch in number_string:
    print ch # will print each digit in order

Or you can slice it:

print number_string[:2] # first two digits
print number_string[-3:] # last three digits
print number_string[3] # forth digit

Or better, don't convert the user's input to an integer (the user types a string)

isbn = raw_input()
for pos, ch in enumerate(reversed(isbn)):
    print "%d * %d is %d" % pos + 2, int(ch), int(ch) * (pos + 2)

For more information read a tutorial.

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雨后咖啡店 2024-07-31 04:30:51

while number:
    digit = number % 10

    # do whatever with digit

    # remove last digit from number (as integer)
    number //= 10

在循环的每次迭代中，它都会从 number 中删除最后一位数字，并将其分配给 digit。
是相反的，从最后一位数字开始，到第一个数字结束

while number:
    digit = number % 10

    # do whatever with digit

    # remove last digit from number (as integer)
    number //= 10

On each iteration of the loop, it removes the last digit from number, assigning it to digit.
It's in reverse, starts from the last digit, finishes with the first

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烈酒灼喉 2024-07-31 04:30:51

list_of_ints = [int(i) for i in str(ISBN)]

会给你一个有序的整数列表。当然，考虑到鸭子类型，您也可以使用 str(ISBN)。

编辑：正如评论中提到的，这个列表不是按升序或降序排序的，但它确实有一个定义的顺序（Python中的集合、字典等理论上没有，尽管实际上顺序倾向于相当可靠）。如果你想对其进行排序：

list_of_ints.sort()

是你的朋友。请注意，sort() 就地排序（实际上更改了现有列表的顺序）并且不返回新列表。

list_of_ints = [int(i) for i in str(ISBN)]

Will give you a ordered list of ints. Of course, given duck typing, you might as well work with str(ISBN).

Edit: As mentioned in the comments, this list isn't sorted in the sense of being ascending or descending, but it does have a defined order (sets, dictionaries, etc in python in theory don't, although in practice the order tends to be fairly reliable). If you want to sort it:

list_of_ints.sort()

is your friend. Note that sort() sorts in place (as in, actually changes the order of the existing list) and doesn't return a new list.

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南笙 2024-07-31 04:30:51

在旧版本的 Python 上...

map(int,str(123))

在新版本 3k 上

list(map(int,str(123)))

On Older versions of Python...

map(int,str(123))

On New Version 3k

list(map(int,str(123)))

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╭ゆ眷念 2024-07-31 04:30:51

(number/10**x)%10

您可以在循环中使用它，其中 number 是完整的数字，x 是循环的每次迭代 (0,1,2,3,...,n)，n 是停止点。 x = 0 给出个位，x = 1 给出十位，x = 2 给出百位，依此类推。请记住，这将给出从右到左的数字值，因此这可能不适用于 ISBN，但它仍然会隔离每个数字。

(number/10**x)%10

You can use this in a loop, where number is the full number, x is each iteration of the loop (0,1,2,3,...,n) with n being the stop point. x = 0 gives the ones place, x = 1 gives the tens, x = 2 gives the hundreds, and so on. Keep in mind that this will give the value of the digits from right to left, so this might not be the for an ISBN but it will still isolate each digit.

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终弃我 2024-07-31 04:30:51

将其转换为字符串并使用 int() 函数对其进行映射。

map(int, str(1231231231))

Convert it to string and map over it with the int() function.

map(int, str(1231231231))

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何处潇湘 2024-07-31 04:30:51

递归版本：

def int_digits(n):
    return [n] if n<10 else int_digits(n/10)+[n%10]

Recursion version:

def int_digits(n):
    return [n] if n<10 else int_digits(n/10)+[n%10]

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执笏见 2024-07-31 04:30:51

转换为 str 肯定比除以 10 慢。

map 比列表理解稍慢：

convert to string with map 2.13599181175
convert to string with list comprehension 1.92812991142
modulo, division, recursive 0.948769807816
modulo, division 0.699964046478

这些时间是由我的笔记本电脑上的以下代码返回的：

foo = """\
def foo(limit):
    return sorted(set(map(sum, map(lambda x: map(int, list(str(x))), map(lambda x: x * 9, range(limit))))))

foo(%i)
"""

bar = """\
def bar(limit):
    return sorted(set([sum([int(i) for i in str(n)]) for n in [k *9 for k in range(limit)]]))

bar(%i)
"""

rac = """\
def digits(n):
    return [n] if n<10 else digits(n / 10)+[n %% 10]

def rabbit(limit):
    return sorted(set([sum(digits(n)) for n in [k *9 for k in range(limit)]]))

rabbit(%i)
"""

rab = """\
def sum_digits(number):
  result = 0
  while number:
    digit = number %% 10
    result += digit
    number /= 10
  return result

def rabbit(limit):
    return sorted(set([sum_digits(n) for n in [k *9 for k in range(limit)]]))

rabbit(%i)
"""


import timeit

print "convert to string with map", timeit.timeit(foo % 100, number=10000)
print "convert to string with list comprehension", timeit.timeit(bar % 100, number=10000)
print "modulo, division, recursive", timeit.timeit(rac % 100, number=10000)
print "modulo, division", timeit.timeit(rab % 100, number=10000)

Converting to str is definitely slower then dividing by 10.

map is sligthly slower than list comprehension:

convert to string with map 2.13599181175
convert to string with list comprehension 1.92812991142
modulo, division, recursive 0.948769807816
modulo, division 0.699964046478

These times were returned by the following code on my laptop:

foo = """\
def foo(limit):
    return sorted(set(map(sum, map(lambda x: map(int, list(str(x))), map(lambda x: x * 9, range(limit))))))

foo(%i)
"""

bar = """\
def bar(limit):
    return sorted(set([sum([int(i) for i in str(n)]) for n in [k *9 for k in range(limit)]]))

bar(%i)
"""

rac = """\
def digits(n):
    return [n] if n<10 else digits(n / 10)+[n %% 10]

def rabbit(limit):
    return sorted(set([sum(digits(n)) for n in [k *9 for k in range(limit)]]))

rabbit(%i)
"""

rab = """\
def sum_digits(number):
  result = 0
  while number:
    digit = number %% 10
    result += digit
    number /= 10
  return result

def rabbit(limit):
    return sorted(set([sum_digits(n) for n in [k *9 for k in range(limit)]]))

rabbit(%i)
"""


import timeit

print "convert to string with map", timeit.timeit(foo % 100, number=10000)
print "convert to string with list comprehension", timeit.timeit(bar % 100, number=10000)
print "modulo, division, recursive", timeit.timeit(rac % 100, number=10000)
print "modulo, division", timeit.timeit(rab % 100, number=10000)

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阳光的暖冬 2024-07-31 04:30:51

经过自己的努力搜索，我发现了几种解决方案，每种方案都有优点和缺点。使用最适合您的任务的。

所有示例均在 GNU/Linux Debian 8 操作系统上使用 CPython 3.5 进行测试。

使用递归

代码

def get_digits_from_left_to_right(number, lst=None):
    """Return digits of an integer excluding the sign."""

    if lst is None:
        lst = list()

    number = abs(number)

    if number < 10:
        lst.append(number)
        return tuple(lst)

    get_digits_from_left_to_right(number // 10, lst)
    lst.append(number % 10)

    return tuple(lst)

演示

In [121]: get_digits_from_left_to_right(-64517643246567536423)
Out[121]: (6, 4, 5, 1, 7, 6, 4, 3, 2, 4, 6, 5, 6, 7, 5, 3, 6, 4, 2, 3)

In [122]: get_digits_from_left_to_right(0)
Out[122]: (0,)

In [123]: get_digits_from_left_to_right(123012312312321312312312)
Out[123]: (1, 2, 3, 0, 1, 2, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 1, 2)

使用函数 divmod

代码

def get_digits_from_right_to_left(number):
    """Return digits of an integer excluding the sign."""

    number = abs(number)

    if number < 10:
        return (number, )

    lst = list()

    while number:
        number, digit = divmod(number, 10)
        lst.insert(0, digit)

    return tuple(lst)

演示

In [125]: get_digits_from_right_to_left(-3245214012321021213)
Out[125]: (3, 2, 4, 5, 2, 1, 4, 0, 1, 2, 3, 2, 1, 0, 2, 1, 2, 1, 3)

In [126]: get_digits_from_right_to_left(0)
Out[126]: (0,)

In [127]: get_digits_from_right_to_left(9999999999999999)
Out[127]: (9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9)

< strong>使用构造tuple(map(int, str(abs(number)))

In [109]: tuple(map(int, str(abs(-123123123))))
Out[109]: (1, 2, 3, 1, 2, 3, 1, 2, 3)

In [110]: tuple(map(int, str(abs(1412421321312))))
Out[110]: (1, 4, 1, 2, 4, 2, 1, 3, 2, 1, 3, 1, 2)

In [111]: tuple(map(int, str(abs(0))))
Out[111]: (0,)

使用函数re.findall

In [112]: tuple(map(int, re.findall(r'\d', str(1321321312))))
Out[112]: (1, 3, 2, 1, 3, 2, 1, 3, 1, 2)

In [113]: tuple(map(int, re.findall(r'\d', str(-1321321312))))
Out[113]: (1, 3, 2, 1, 3, 2, 1, 3, 1, 2)

In [114]: tuple(map(int, re.findall(r'\d', str(0))))
Out[114]: (0,)

使用模块decimal

In [117]: decimal.Decimal(0).as_tuple().digits
Out[117]: (0,)

In [118]: decimal.Decimal(3441120391321).as_tuple().digits
Out[118]: (3, 4, 4, 1, 1, 2, 0, 3, 9, 1, 3, 2, 1)

In [119]: decimal.Decimal(-3441120391321).as_tuple().digits
Out[119]: (3, 4, 4, 1, 1, 2, 0, 3, 9, 1, 3, 2, 1)

After own diligent searches I found several solutions, where each has advantages and disadvantages. Use the most suitable for your task.

All examples tested with the CPython 3.5 on the operation system GNU/Linux Debian 8.

Using a recursion

Code

def get_digits_from_left_to_right(number, lst=None):
    """Return digits of an integer excluding the sign."""

    if lst is None:
        lst = list()

    number = abs(number)

    if number < 10:
        lst.append(number)
        return tuple(lst)

    get_digits_from_left_to_right(number // 10, lst)
    lst.append(number % 10)

    return tuple(lst)

Demo

In [121]: get_digits_from_left_to_right(-64517643246567536423)
Out[121]: (6, 4, 5, 1, 7, 6, 4, 3, 2, 4, 6, 5, 6, 7, 5, 3, 6, 4, 2, 3)

In [122]: get_digits_from_left_to_right(0)
Out[122]: (0,)

In [123]: get_digits_from_left_to_right(123012312312321312312312)
Out[123]: (1, 2, 3, 0, 1, 2, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 1, 2)

Using the function divmod

Code

def get_digits_from_right_to_left(number):
    """Return digits of an integer excluding the sign."""

    number = abs(number)

    if number < 10:
        return (number, )

    lst = list()

    while number:
        number, digit = divmod(number, 10)
        lst.insert(0, digit)

    return tuple(lst)

Demo

In [125]: get_digits_from_right_to_left(-3245214012321021213)
Out[125]: (3, 2, 4, 5, 2, 1, 4, 0, 1, 2, 3, 2, 1, 0, 2, 1, 2, 1, 3)

In [126]: get_digits_from_right_to_left(0)
Out[126]: (0,)

In [127]: get_digits_from_right_to_left(9999999999999999)
Out[127]: (9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9)

Using a construction tuple(map(int, str(abs(number)))

In [109]: tuple(map(int, str(abs(-123123123))))
Out[109]: (1, 2, 3, 1, 2, 3, 1, 2, 3)

In [110]: tuple(map(int, str(abs(1412421321312))))
Out[110]: (1, 4, 1, 2, 4, 2, 1, 3, 2, 1, 3, 1, 2)

In [111]: tuple(map(int, str(abs(0))))
Out[111]: (0,)

Using the function re.findall

In [112]: tuple(map(int, re.findall(r'\d', str(1321321312))))
Out[112]: (1, 3, 2, 1, 3, 2, 1, 3, 1, 2)

In [113]: tuple(map(int, re.findall(r'\d', str(-1321321312))))
Out[113]: (1, 3, 2, 1, 3, 2, 1, 3, 1, 2)

In [114]: tuple(map(int, re.findall(r'\d', str(0))))
Out[114]: (0,)

Using the module decimal

In [117]: decimal.Decimal(0).as_tuple().digits
Out[117]: (0,)

In [118]: decimal.Decimal(3441120391321).as_tuple().digits
Out[118]: (3, 4, 4, 1, 1, 2, 0, 3, 9, 1, 3, 2, 1)

In [119]: decimal.Decimal(-3441120391321).as_tuple().digits
Out[119]: (3, 4, 4, 1, 1, 2, 0, 3, 9, 1, 3, 2, 1)

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好倦 2024-07-31 04:30:51

使用此循环的主体对数字执行任何您想要的操作

for digit in map(int, str(my_number)):

Use the body of this loop to do whatever you want to with the digits

for digit in map(int, str(my_number)):

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故事和酒 2024-07-31 04:30:51

我编写了这个程序，这是在我的程序中实际计算校验位的代码

    #Get the 10 digit number
    number=input("Please enter ISBN number: ")

    #Explained below
    no11 = (((int(number[0])*11) + (int(number[1])*10) + (int(number[2])*9) + (int(number[3])*8) 
           + (int(number[4])*7) + (int(number[5])*6) + (int(number[6])*5) + (int(number[7])*4) +
           (int(number[8])*3) + (int(number[9])*2))/11)

    #Round to 1 dp
    no11 = round(no11, 1)

    #explained below
    no11 = str(no11).split(".")

    #get the remainder and check digit
    remainder = no11[1]
    no11 = (11 - int(remainder))

    #Calculate 11 digit ISBN
    print("Correct ISBN number is " + number + str(no11))

，它是一长行代码，但它将数字分开，将数字乘以适当的数量，将它们加在一起并除以它们11、一行代码。 .split() 函数只是创建一个列表（按小数点拆分），因此您可以获取列表中的第二项，然后从 11 中获取校验位。通过更改这两行也可以提高效率：

    remainder = no11[1]
    no11 = (11 - int(remainder))

为此：

    no11 = (11 - int(no11[1]))

希望这有帮助:)

I have made this program and here is the bit of code that actually calculates the check digit in my program

    #Get the 10 digit number
    number=input("Please enter ISBN number: ")

    #Explained below
    no11 = (((int(number[0])*11) + (int(number[1])*10) + (int(number[2])*9) + (int(number[3])*8) 
           + (int(number[4])*7) + (int(number[5])*6) + (int(number[6])*5) + (int(number[7])*4) +
           (int(number[8])*3) + (int(number[9])*2))/11)

    #Round to 1 dp
    no11 = round(no11, 1)

    #explained below
    no11 = str(no11).split(".")

    #get the remainder and check digit
    remainder = no11[1]
    no11 = (11 - int(remainder))

    #Calculate 11 digit ISBN
    print("Correct ISBN number is " + number + str(no11))

Its a long line of code, but it splits the number up, multiplies the digits by the appropriate amount, adds them together and divides them by 11, in one line of code. The .split() function just creates a list (being split at the decimal) so you can take the 2nd item in the list and take that from 11 to find the check digit. This could also be made even more efficient by changing these two lines:

    remainder = no11[1]
    no11 = (11 - int(remainder))

To this:

    no11 = (11 - int(no11[1]))

Hope this helps :)

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南冥有猫 2024-07-31 04:30:51

类似于 this 答案，但更“Pythonic”的方式迭代数字将是：

while number:
    # "pop" the rightmost digit
    number, digit = divmod(number, 10)

Similar to this answer but more a more "pythonic" way to iterate over the digis would be:

while number:
    # "pop" the rightmost digit
    number, digit = divmod(number, 10)

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糖果控 2024-07-31 04:30:51

一行数字列表怎么样？

ldigits = lambda n, l=[]: not n and l or l.insert(0,n%10) or ldigits(n/10,l)

How about a one-liner list of digits...

ldigits = lambda n, l=[]: not n and l or l.insert(0,n%10) or ldigits(n/10,l)

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苍暮颜 2024-07-31 04:30:51

答案： 165

方法：暴力破解！这里有一小段 Python（2.7 版）代码来统计所有数据。

from math import sqrt, floor
is_ps = lambda x: floor(sqrt(x)) ** 2 == x
count = 0
for n in range(1002, 10000, 3):
    if n % 11 and is_ps(sum(map(int, str(n)))):
        count += 1
        print "#%i: %s" % (count, n)

Answer: 165

Method: brute-force! Here is a tiny bit of Python (version 2.7) code to count'em all.

from math import sqrt, floor
is_ps = lambda x: floor(sqrt(x)) ** 2 == x
count = 0
for n in range(1002, 10000, 3):
    if n % 11 and is_ps(sum(map(int, str(n)))):
        count += 1
        print "#%i: %s" % (count, n)

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标点 2024-07-31 04:30:51

假设您想从整数 x 中获取第 i 个最低有效数字，您可以尝试：

(abs(x)%(10**i))/(10**(i-1))

我希望它有所帮助。

Just assuming you want to get the i-th least significant digit from an integer number x, you can try:

(abs(x)%(10**i))/(10**(i-1))

I hope it helps.

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