假设我有一个输入整数
12345
。我怎样才能将它分成像[1, 2, 3, 4, 5]
这样的列表?
>>> list(str(12345))
['1', '2', '3', '4', '5']
以整数形式返回数组
>>> map(int,str(12345))
[1, 2, 3, 4, 5]
def digitize(n, base=10):
if n == 0:
yield 0
while n:
n, d = divmod(n, base)
yield d
示例:
tuple(digitize(123456789)) == (9, 8, 7, 6, 5, 4, 3, 2, 1)
tuple(digitize(0b1101110, 2)) == (0, 1, 1, 1, 0, 1, 1)
tuple(digitize(0x123456789ABCDEF, 16)) == (15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
如您所见,这将产生从右到左的数字。如果您想要从左到右的数字,您需要创建一个序列,然后反转它:
reversed(tuple(digitize(x)))
您还可以在分割整数时使用此函数进行基数转换。以下示例将十六进制数拆分为二进制半字节作为元组:
import itertools as it
tuple(it.zip_longest(*[digitize(0x123456789ABCDEF, 2)]*4, fillvalue=0)) == ((1, 1, 1, 1), (0, 1, 1, 1), (1, 0, 1, 1), (0, 0, 1, 1), (1, 1, 0, 1), (0, 1, 0, 1), (1, 0, 0, 1), (0, 0, 0, 1), (1, 1, 1, 0), (0, 1, 1, 0), (1, 0, 1, 0), (0, 0, 1, 0), (1, 1, 0, 0), (0, 1, 0, 0), (1, 0, 0, 0))
请注意,此方法不处理小数,但可以进行调整。
[int(i) for i in str(number)]
或者,如果不想使用列表理解或者您想使用不同于 10 的基数
from __future__ import division # for compatibility of // between Python 2 and 3
def digits(number, base=10):
assert number >= 0
if number == 0:
return [0]
l = []
while number > 0:
l.append(number % base)
number = number // base
return l
list(map(int, str(x)))
是 Pythonic 方法,但您可以制定逻辑来派生数字而无需任何类型转换:
from math import log10
def digitize(x):
n = int(log10(x))
for i in range(n, -1, -1):
factor = 10**i
k = x // factor
yield k
x -= k * factor
res = list(digitize(5243))
[5, 2, 4, 3]
生成器的一个好处是您可以无缝地提供
set
、
tuple
、
next
等,无需任何额外的逻辑。
>>> [ int(i,16) for i in '0123456789ABCDEF' ]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
>>> [int(i,2) for i in "100 010 110 111".split()]
[4, 2, 6, 7]
from math import log10
def decompose(n):
if n == 0:
return [0]
b = int(log10(n)) + 1
return [(n // (10 ** i)) % 10 for i in reversed(range(b))]
join
和
split
方法:
>>> a=12345
>>> list(map(int,' '.join(str(a)).split()))
[1, 2, 3, 4, 5]
>>> [int(i) for i in ' '.join(str(a)).split()]
[1, 2, 3, 4, 5]
>>>
这里我们还使用 map
或列表理解来获取列表。
str
和列表理解是最具可读性和最快的方法:
def int_to_list():
q = 1111111111111122222222222222222333333333333333333333
res = []
while q != 0:
res.append(q % 10)
q //= 10
return res[::-1]
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))
0.1116315030012629
def int_to_list():
q = 1111111111111122222222222222222333333333333333333333
return [int(x) for x in str(q)]
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))
0.10781958399820724
str(12345)
nums = []
c = 12345
for i in str(c):
l = i.split()[0]
nums.append(l)
np.array(nums)