python将整数变成列表_python怎么把整数变成列表_如何将多个整数的列表转换为单个整数？...

仅使用数学(不转换为字符串或从字符串转换为字符串)，可以使用reduce函数(Python 3中的functools.reduce)b = reduce(lambda total, d: 10*total + d, x, 0)

这利用了Horner法则，它将表示该数的多项式进行因子化以减少乘法数。例如1357 = 1*10*10*10 + 3*10*10 + 5*10 + 7 # 6 multiplications

= ((1*10 + 3)*10 + 5)*10 + 7 # 3 multiplications

因此，这比计算能力10或创建字符串并将结果转换为整数更快。>>> timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', 'from functools import reduce; x=[1,3,5,7]')

0.7217515400843695

>>> timeit.timeit('int("".join(map(str, [1,3,5,7])))')

1.425914661027491

>>> timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', 'x=[1,3,5,7]')

1.897974518011324

公平地说，一旦位数变大，字符串转换就会更快。>>> import timeit

# 30 digits

>>> setup='from functools import reduce; x=[5, 2, 6, 8, 4, 6, 6, 4, 8, 0, 3, 1, 7, 6, 8, 2, 9, 9, 9, 5, 4, 5, 5, 4, 3, 6, 9, 2, 2, 1]'

>>> print(timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', setup))

6.520374411018565

>>> print(timeit.timeit('int("".join(map(str, x)))', setup))

6.797425839002244

>>> print(timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', setup))

19.430233853985555

# 60 digits

>>> setup='from functools import reduce; x=2*[5, 2, 6, 8, 4, 6, 6, 4, 8, 0, 3, 1, 7, 6, 8, 2, 9, 9, 9, 5, 4, 5, 5, 4, 3, 6, 9, 2, 2, 1]'

>>> print(timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', setup))

13.648188541992567

>>> print(timeit.timeit('int("".join(map(str, x)))', setup))

12.864593736943789

>>> print(timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', setup))

44.141602706047706

# 120 digits!

>>> setup='from functools import reduce; x=4*[5, 2, 6, 8, 4, 6, 6, 4, 8, 0, 3, 1, 7, 6, 8, 2, 9, 9, 9, 5, 4, 5, 5, 4, 3, 6, 9, 2, 2, 1]'

>>> print(timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', setup))

28.364255172084086

>>> print(timeit.timeit('int("".join(map(str, x)))', setup))

25.184791765059344

>>> print(timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', setup))

99.88558598596137