将整数列表转换为一个数字？

我有一个整数列表,我想转换为一个数字,如:

numList = [1, 2, 3]
num = magic(numList)

print num, type(num)
>>> 123, 

实现魔术功能的最佳方法是什么？

编辑
我确实找到了这个,但似乎必须有一个更好的方法.

1> Triptych..：

---

# Over-explaining a bit:
def magic(numList):         # [1,2,3]
    s = map(str, numList)   # ['1','2','3']
    s = ''.join(s)          # '123'
    s = int(s)              # 123
    return s


# How I'd probably write it:
def magic(numList):
    s = ''.join(map(str, numList))
    return int(s)


# As a one-liner  
num = int(''.join(map(str,numList)))


# Functionally:
s = reduce(lambda x,y: x+str(y), numList, '')
num = int(s)


# Using some oft-forgotten built-ins:
s = filter(str.isdigit, repr(numList))
num = int(s)

---

TokenMacGuy:你的意思是这个？ - http://www.artima.com/weblogs/viewpost.jsp?thread=98196 map,reduce,filter,lambda原来都是3k

---

函数方法的小改进:reduce(lambda x,y:10*x + y,numList)

2> cdleary..：

---

两种解决方案

>>> nums = [1, 2, 3]
>>> magic = lambda nums: int(''.join(str(i) for i in nums)) # Generator exp.
>>> magic(nums)
123
>>> magic = lambda nums: sum(digit * 10 ** (len(nums) - 1 - i) # Summation
...     for i, digit in enumerate(nums))
>>> magic(nums)
123

该map取向的解决方案实际上出来向前在我的箱子-你绝对不应该使用sum的东西,可能是大数:

import collections
import random
import timeit

import matplotlib.pyplot as pyplot

MICROSECONDS_PER_SECOND = 1E6
FUNS = []
def test_fun(fun):
    FUNS.append(fun)
    return fun

@test_fun
def with_map(nums):
    return int(''.join(map(str, nums)))

@test_fun
def with_interpolation(nums):
    return int(''.join('%d' % num for num in nums))

@test_fun
def with_genexp(nums):
    return int(''.join(str(num) for num in nums))

@test_fun
def with_sum(nums):
    return sum(digit * 10 ** (len(nums) - 1 - i)
        for i, digit in enumerate(nums))

@test_fun
def with_reduce(nums):
    return int(reduce(lambda x, y: x + str(y), nums, ''))

@test_fun
def with_builtins(nums):
    return int(filter(str.isdigit, repr(nums)))

@test_fun
def with_accumulator(nums):
    tot = 0
    for num in nums:
        tot *= 10
        tot += num
    return tot

def time_test(digit_count, test_count=10000):
    """
    :return: Map from func name to (normalized) microseconds per pass.
    """
    print 'Digit count:', digit_count
    nums = [random.randrange(1, 10) for i in xrange(digit_count)]
    stmt = 'to_int(%r)' % nums
    result_by_method = {}
    for fun in FUNS:
        setup = 'from %s import %s as to_int' % (__name__, fun.func_name)
        t = timeit.Timer(stmt, setup)
        per_pass = t.timeit(number=test_count) / test_count
        per_pass *= MICROSECONDS_PER_SECOND
        print '%20s: %.2f usec/pass' % (fun.func_name, per_pass)
        result_by_method[fun.func_name] = per_pass
    return result_by_method

if __name__ == '__main__':
    pass_times_by_method = collections.defaultdict(list)
    assert_results = [fun([1, 2, 3]) for fun in FUNS]
    assert all(result == 123 for result in assert_results)
    digit_counts = range(1, 100, 2)
    for digit_count in digit_counts:
        for method, result in time_test(digit_count).iteritems():
            pass_times_by_method[method].append(result)
    for method, pass_times in pass_times_by_method.iteritems():
        pyplot.plot(digit_counts, pass_times, label=method)
    pyplot.legend(loc='upper left')
    pyplot.xlabel('Number of Digits')
    pyplot.ylabel('Microseconds')
    pyplot.show()

3> Rex Logan..：

---

def magic(number):
    return int(''.join(str(i) for i in number))

4> TokenMacGuy..：

---

def magic(numbers):
    return int(''.join([ "%d"%x for x in numbers]))

5> jfs..：

---

为了完整print()起见,这是一个使用的变体(适用于Python 2.6-3.x):

from __future__ import print_function
try: from cStringIO import StringIO
except ImportError:
     from io import StringIO

def to_int(nums, _s = StringIO()):
    print(*nums, sep='', end='', file=_s)
    s = _s.getvalue()
    _s.truncate(0)
    return int(s)

不同解决方案的时间表现

我已经测量了@cdleary函数的性能.结果略有不同.

每个函数都使用以下生成的输入列表进行测试

def randrange1_10(digit_count): # same as @cdleary
    return [random.randrange(1, 10) for i in xrange(digit_count)]

您可以通过--sequence-creator=yourmodule.yourfunction命令行参数提供自己的函数(参见下文).

列表(len(nums) == digit_count)中给定数量的整数的最快函数是:

len(nums)在1..30

def _accumulator(nums):
    tot = 0
    for num in nums:
        tot *= 10
        tot += num
    return tot

len(nums)在30..1000

def _map(nums):
    return int(''.join(map(str, nums)))

def _imap(nums):
    return int(''.join(imap(str, nums)))

|------------------------------+-------------------|
| Fitting polynom              | Function          |
|------------------------------+-------------------|
| 1.00  log2(N)   +  1.25e-015 | N                 |
| 2.00  log2(N)   +  5.31e-018 | N*N               |
| 1.19  log2(N)   +      1.116 | N*log2(N)         |
| 1.37  log2(N)   +      2.232 | N*log2(N)*log2(N) |
|------------------------------+-------------------|
| 1.21  log2(N)   +      0.063 | _interpolation    |
| 1.24  log2(N)   -      0.610 | _genexp           |
| 1.25  log2(N)   -      0.968 | _imap             |
| 1.30  log2(N)   -      1.917 | _map              |

要绘制第一个图形下载cdleary.py并make-figures.py运行(numpy并且matplotlib必须安装到绘图中):

$ python cdleary.py 

要么

$ python make-figures.py --sort-function=cdleary._map \
> --sort-function=cdleary._imap \
> --sort-function=cdleary._interpolation \
> --sort-function=cdleary._genexp --sort-function=cdleary._sum \
> --sort-function=cdleary._reduce --sort-function=cdleary._builtins \
> --sort-function=cdleary._accumulator \
> --sequence-creator=cdleary.randrange1_10 --maxn=1000