Python允许从给定基数的字符串中轻松创建一个整数
int(str, base).
我想执行相反的操作:从一个整数创建一个字符串， 例如，我想要一些函数int2base(num, base)，这样:
int(int2base(x, b), b) == x
函数名/参数的顺序并不重要。
对于int()将接受的任何以b为底的数字x。
这是一个很容易写的函数:事实上，它比在这个问题中描述它更容易。然而，我觉得我一定是错过了什么。
我知道函数bin, oct, hex，但我不能使用它们的几个原因:
这些函数在旧版本的Python中不可用，我需要与(2.2)兼容 我想要一个通解对于不同的碱都可以用同样的方式表示 我想允许2 8 16以外的底数
相关的
Python优雅的int(string, base)逆函数 在python中使用递归的整数到base-x系统 Python中的Base 62转换 如何在Python中将整数转换为最短的url安全字符串?
当前回答
>>> numpy.base_repr(10, base=3)
'101'
注意，numpy.base_repr()的基数限制为36。否则抛出ValueError
其他回答
递归
我将投票最多的答案简化为:
BS="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def to_base(n, b):
return "0" if not n else to_base(n//b, b).lstrip("0") + BS[n%b]
对于RuntimeError有相同的建议:对于非常大的整数和负数，在cmp中超过最大递归深度。(你可以使用setrecursionlimit(new_limit))
迭代
为了避免递归问题:
BS="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def to_base(s, b):
res = ""
while s:
res+=BS[s%b]
s//= b
return res[::-1] or "0"
假设我们想把14转换成2进制。我们反复应用除法算法，直到商为0:
14 = 2 × 7
7 = 2 × 3 + 1
3 = 2 × 1 + 1
1 = 2 × 0 + 1
二进制表示就是从下往上读的余数。这可以通过展开来证明
14 = 2 × 7 = 2 × (2 × 3 + 1) = 2 × (2 × (2 × 1 + 1) + 1) = 2 × (2 × (2 × 0 + 1) + 1) = 2^3 + 2^2 + 2
本代码是上述算法的实现。
def toBaseX(n, X):
strbin = ""
while n != 0:
strbin += str(n % X)
n = n // X
return strbin[::-1]
def int2base(a, base, numerals="0123456789abcdefghijklmnopqrstuvwxyz"):
baseit = lambda a=a, b=base: (not a) and numerals[0] or baseit(a-a%b,b*base)+numerals[a%b%(base-1) or (a%b) and (base-1)]
return baseit()
解释
在任何底数下，每个数字都等于a1+a2*base**2+a3*base**3…“任务”是找出所有的a。
everyN = 1、2、3……代码通过b对b=base**(N+1)进行“模组”来隔离aN*base**N, b=base**(N+1)切片所有大于N的a，并通过每次由当前aN*base**N调用func时减少a来切片它们的序列小于N的所有a。
底%(底-1)==1，则底**p%(底-1)==1，而底q*底^p%(底-1)==q，只有当q=底-1时例外，返回0。 为了解决这个问题，如果它返回0,func会检查它从原点开始是否是0。
优势
在这个例子中，只有一个乘法(而不是除法)和一些模量运算，这些运算相对花费的时间较少。
The below provided Python code converts a Python integer to a string in arbitrary base ( from 2 up to infinity ) and works in both directions. So all the created strings can be converted back to Python integers by providing a string for N instead of an integer. The code works only on positive numbers by intention (there is in my eyes some hassle about negative values and their bit representations I don't want to dig into). Just pick from this code what you need, want or like, or just have fun learning about available options. Much is there only for the purpose of documenting all the various available approaches ( e.g. the Oneliner seems not to be fast, even if promised to be ).
我喜欢萨尔瓦多·达利提出的无限大基地的格式。一个很好的建议，它在光学上工作得很好，即使是简单的二进制位表示。注意，在infiniteBase=True格式的字符串的情况下，width=x填充参数适用于数字，而不是整个数字。似乎，代码处理无穷大数字格式运行甚至比其他选项快一点-使用它的另一个原因?
我不喜欢使用Unicode来扩展数字可用的符号数量的想法，所以不要在下面的代码中寻找它，因为它不存在。请使用建议的infiniteBase格式，或者将整数存储为字节以进行压缩。
def inumToStr( N, base=2, width=1, infiniteBase=False,\
useNumpy=False, useRecursion=False, useOneliner=False, \
useGmpy=False, verbose=True):
''' Positive numbers only, but works in BOTH directions.
For strings in infiniteBase notation set for bases <= 62
infiniteBase=True . Examples of use:
inumToStr( 17, 2, 1, 1) # [1,0,0,0,1]
inumToStr( 17, 3, 5) # 00122
inumToStr(245, 16, 4) # 00F5
inumToStr(245, 36, 4,0,1) # 006T
inumToStr(245245245245,36,10,0,1) # 0034NWOQBH
inumToStr(245245245245,62) # 4JhA3Th
245245245245 == int(gmpy2.mpz('4JhA3Th',62))
inumToStr(245245245245,99,2) # [25,78, 5,23,70,44]
----------------------------------------------------
inumToStr( '[1,0,0,0,1]',2, infiniteBase=True ) # 17
inumToStr( '[25,78, 5,23,70,44]', 99) # 245245245245
inumToStr( '0034NWOQBH', 36 ) # 245245245245
inumToStr( '4JhA3Th' , 62 ) # 245245245245
----------------------------------------------------
--- Timings for N = 2**4096, base=36:
standard: 0.0023
infinite: 0.0017
numpy : 0.1277
recursio; 0.0022
oneliner: 0.0146
For N = 2**8192:
standard: 0.0075
infinite: 0.0053
numpy : 0.1369
max. recursion depth exceeded: recursio/oneliner
'''
show = print
if type(N) is str and ( infiniteBase is True or base > 62 ):
lstN = eval(N)
if verbose: show(' converting a non-standard infiniteBase bits string to Python integer')
return sum( [ item*base**pow for pow, item in enumerate(lstN[::-1]) ] )
if type(N) is str and base <= 36:
if verbose: show('base <= 36. Returning Python int(N, base)')
return int(N, base)
if type(N) is str and base <= 62:
if useGmpy:
if verbose: show(' base <= 62, useGmpy=True, returning int(gmpy2.mpz(N,base))')
return int(gmpy2.mpz(N,base))
else:
if verbose: show(' base <= 62, useGmpy=False, self-calculating return value)')
lstStrOfDigits="0123456789"+ \
"abcdefghijklmnopqrstuvwxyz".upper() + \
"abcdefghijklmnopqrstuvwxyz"
dictCharToPow = {}
for index, char in enumerate(lstStrOfDigits):
dictCharToPow.update({char : index})
return sum( dictCharToPow[item]*base**pow for pow, item in enumerate(N[::-1]) )
#:if
#:if
if useOneliner and base <= 36:
if verbose: show(' base <= 36, useOneliner=True, running the Oneliner code')
d="0123456789abcdefghijklmnopqrstuvwxyz"
baseit = lambda a=N, b=base: (not a) and d[0] or \
baseit(a-a%b,b*base)+d[a%b%(base-1) or (a%b) and (base-1)]
return baseit().rjust(width, d[0])[1:]
if useRecursion and base <= 36:
if verbose: show(' base <= 36, useRecursion=True, running recursion algorythm')
BS="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def to_base(n, b):
return "0" if not n else to_base(n//b, b).lstrip("0") + BS[n%b]
return to_base(N, base).rjust(width,BS[0])
if base > 62 or infiniteBase:
if verbose: show(' base > 62 or infiniteBase=True, returning a non-standard digits string')
# Allows arbitrary large base with 'width=...'
# applied to each digit (useful also for bits )
N, digit = divmod(N, base)
strN = str(digit).rjust(width, ' ')+']'
while N:
N, digit = divmod(N, base)
strN = str(digit).rjust(width, ' ') + ',' + strN
return '[' + strN
#:if
if base == 2:
if verbose: show(" base = 2, returning Python str(f'{N:0{width}b}')")
return str(f'{N:0{width}b}')
if base == 8:
if verbose: show(" base = 8, returning Python str(f'{N:0{width}o}')")
return str(f'{N:0{width}o}')
if base == 16:
if verbose: show(" base = 16, returning Python str(f'{N:0{width}X}')")
return str(f'{N:0{width}X}')
if base <= 36:
if useNumpy:
if verbose: show(" base <= 36, useNumpy=True, returning np.base_repr(N, base)")
import numpy as np
strN = np.base_repr(N, base)
return strN.rjust(width, '0')
else:
if verbose: show(' base <= 36, useNumpy=False, self-calculating return value)')
lstStrOfDigits="0123456789"+"abcdefghijklmnopqrstuvwxyz".upper()
strN = lstStrOfDigits[N % base] # rightmost digit
while N >= base:
N //= base # consume already converted digit
strN = lstStrOfDigits[N % base] + strN # add digits to the left
#:while
return strN.rjust(width, lstStrOfDigits[0])
#:if
#:if
if base <= 62:
if useGmpy:
if verbose: show(" base <= 62, useGmpy=True, returning gmpy2.digits(N, base)")
import gmpy2
strN = gmpy2.digits(N, base)
return strN.rjust(width, '0')
# back to Python int from gmpy2.mpz with
# int(gmpy2.mpz('4JhA3Th',62))
else:
if verbose: show(' base <= 62, useGmpy=False, self-calculating return value)')
lstStrOfDigits= "0123456789" + \
"abcdefghijklmnopqrstuvwxyz".upper() + \
"abcdefghijklmnopqrstuvwxyz"
strN = lstStrOfDigits[N % base] # rightmost digit
while N >= base:
N //= base # consume already converted digit
strN = lstStrOfDigits[N % base] + strN # add digits to the left
#:while
return strN.rjust(width, lstStrOfDigits[0])
#:if
#:if
#:def
很棒的答案! 我想我问题的答案是“不”，我并没有错过一些明显的解决方案。 下面是我将使用的函数，它可以浓缩答案中所表达的好想法。
允许调用者提供的字符映射(允许base64编码) 检查负数和零 将复数映射为字符串元组
def int2base(x,b,alphabet='0123456789abcdefghijklmnopqrstuvwxyz'): 'convert an integer to its string representation in a given base' if b<2 or b>len(alphabet): if b==64: # assume base64 rather than raise error alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/" else: raise AssertionError("int2base base out of range") if isinstance(x,complex): # return a tuple return ( int2base(x.real,b,alphabet) , int2base(x.imag,b,alphabet) ) if x<=0: if x==0: return alphabet[0] else: return '-' + int2base(-x,b,alphabet) # else x is non-negative real rets='' while x>0: x,idx = divmod(x,b) rets = alphabet[idx] + rets return rets
