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
我写了这个函数，我用它来编码不同的碱基。我还提供了通过值“offset”来移动结果的方法。如果你想编码到64进制以上，但保持可显示字符(如95进制)，这是有用的。
我还试图避免反转输出“列表”，并尽量减少计算操作。pow(base)数组是根据需要计算的，并保留用于对函数的其他调用。
输出是一个二进制字符串
pows = {}
######################################################
def encode_base(value,
base = 10,
offset = 0) :
"""
Encode value into a binary string, according to the desired base.
Input :
value : Any positive integer value
offset : Shift the encoding (eg : Starting at chr(32))
base : The base in which we'd like to encode the value
Return : Binary string
Example : with : offset = 32, base = 64
100 -> !D
200 -> #(
"""
# Determine the number of loops
try :
pb = pows[base]
except KeyError :
pb = pows[base] = {n : base ** n for n in range(0, 8) if n < 2 ** 48 -1}
for n in pb :
if value < pb[n] :
n -= 1
break
out = []
while n + 1 :
b = pb[n]
out.append(chr(offset + value // b))
n -= 1
value %= b
return ''.join(out).encode()
其他回答
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
如果你需要兼容Python的古老版本，你可以使用gmpy(它包含一个快速的，完全通用的int-to-string转换函数，可以为这样的古老版本构建-你可能需要尝试更老的版本，因为最近的版本还没有针对古老的Python和GMP版本进行测试，只有一些最近的版本)，或者，为了速度较慢但更方便，使用Python代码-例如，对于Python 2，最简单的方法是:
import string
digs = string.digits + string.ascii_letters
def int2base(x, base):
if x < 0:
sign = -1
elif x == 0:
return digs[0]
else:
sign = 1
x *= sign
digits = []
while x:
digits.append(digs[int(x % base)])
x = int(x / base)
if sign < 0:
digits.append('-')
digits.reverse()
return ''.join(digits)
对于Python 3, int(x / base)会导致不正确的结果，必须将其更改为x // base:
import string
digs = string.digits + string.ascii_letters
def int2base(x, base):
if x < 0:
sign = -1
elif x == 0:
return digs[0]
else:
sign = 1
x *= sign
digits = []
while x:
digits.append(digs[x % base])
x = x // base
if sign < 0:
digits.append('-')
digits.reverse()
return ''.join(digits)
num = input("number")
power = 0
num = int(num)
while num > 10:
num = num / 10
power += 1
print(str(round(num, 2)) + "^" + str(power))
def baseConverter(x, b):
s = ""
d = string.printable.upper()
while x > 0:
s += d[x%b]
x = x / b
return s[::-1]
字符串不是表示数字的唯一选择:您可以使用一个整数列表来表示每个数字的顺序。这些可以很容易地转换为字符串。
没有一个答案拒绝底数< 2;对于非常大的数字(如56789 ** 43210)，大多数将运行非常缓慢或因堆栈溢出而崩溃。为了避免这种失败，可以像这样快速减少:
def n_to_base(n, b):
if b < 2: raise # invalid base
if abs(n) < b: return [n]
ret = [y for d in n_to_base(n, b*b) for y in divmod(d, b)]
return ret[1:] if ret[0] == 0 else ret # remove leading zeros
def base_to_n(v, b):
h = len(v) // 2
if h == 0: return v[0]
return base_to_n(v[:-h], b) * (b**h) + base_to_n(v[-h:], b)
assert ''.join(['0123456789'[x] for x in n_to_base(56789**43210,10)])==str(56789**43210)
在速度方面，n_to_base对于较大的数字(在我的机器上约为0.3秒)与str相当，但如果与十六进制进行比较，您可能会感到惊讶(在我的机器上约为0.3毫秒，或快1000倍)。这是因为大整数以256(字节)为基数存储在内存中。每个字节可以简单地转换为两个字符的十六进制字符串。这种对齐只发生在底数为2的幂的情况下，这就是为什么有2、8和16(以及base64, ascii, utf16, utf32)的特殊情况。
Consider the last digit of a decimal string. How does it relate to the sequence of bytes that forms its integer? Let's label the bytes s[i] with s[0] being the least significant (little endian). Then the last digit is sum([s[i]*(256**i) % 10 for i in range(n)]). Well, it happens that 256**i ends with a 6 for i > 0 (6*6=36) so that last digit is (s[0]*5 + sum(s)*6)%10. From this, you can see that the last digit depends on the sum of all the bytes. This nonlocal property is what makes converting to decimal harder.
