Question

路线：

正数y，lg y的二进制对数是指数   x，y == 2 ** x。例如，自lg 8 = 3以来8 == 2 ** 3，   自lg(1024) = 10，1024 == 2 ** 10和lg(1025) ≈ 10.0014以来lg(2013.15) ≈ 10.9752 {。}}   lgCeil(x)。

实现x，它返回二进制日志的上限   x ≥ 1 math。请勿导入while模块。相反，使用   lgCeil(8)循环。 3返回lgCeil(1024); 10返回lgCeil(1025)，   lgCeil(2013.15)和11都返回def lgCeil(x): startNum = 0 numOfTimes = 0 base = 2 integer = base * 2 while not(integer >= x): startNum = 2 numOfTimes = numOfTimes + 1 integer = integer * base numOfTimes = numOfTimes + startNum if (integer <= x): base**(numOfTimes+1) return numOfTimes print (lgCeil(8)) print (lgCeil(1024)) print (lgCeil(1025)) print (lgCeil(2013.15)) print (lgCeil(4))。

我的代码：

{{1}}

我不确定我需要修理什么。

Answer 1

你应该用值而不是上升来降低到零。我写了一个工作样本，但它使用了移位，所以你的教授很可能你没有写它。你不应该简单地复制它！尝试理解并重现它！

# This is not Java, functions are lowercase with underscore.
def lg_ceil(value):
    '''
    calculate the n for 2^n <= value < 2^(n + 1) with a while loop (actually a bad idea)
    '''
    # make sure we have an int
    value = int(value)
    # not defined for values <= 1
    if value <= 1:
        return
    result = 0

    # as long as value != 0
    while value:
        # shift value one bit, this equals / 2 with integers.
        value = value >> 1
        # count the numbers of shifts
        result += 1
    # the result is the number shifts minus 1.
    return result - 1

Answer 2

这比你做的更简单更简单 - 你有大致正确的实现（增加一个数字，直到你等于或超过输入），但有很多破坏事物的复杂性。

def lgCeil(x):
    startNum = 0
    numOfTimes = 0 # why two values to keep count?
    base = 2
    integer = base * 2 # what about numbers <= 2?
    while not(integer >= x): # why not 'integer < x'?
        startNum = 2 # always resets to 2 - why?
        numOfTimes = numOfTimes + 1 # could just be '+= 1'
        integer = integer * base # could just be '*= base'
    numOfTimes = numOfTimes + startNum # always adding on a fixed 2
    if (integer <= x):
        base**(numOfTimes+1) # this line doesn't actually contribute to output
    return numOfTimes

考虑这个更简单的实现：

def log_ceil(num, base=2):
    """Calculate the 'ceiling log' for the specified number."""
    val = 0
    while (base ** val) < num:
        val += 1
    return val

这也符合style guide。

Answer 3

def lgCeil(x):
    numOfTimes = 0
    base=2
    integer = 2
    while not(integer >= x):
        numOfTimes = numOfTimes + 1
        integer = integer * base
    return numOfTimes+1

Answer 4

正如jonrsharpe指出的那样，你的代码最小化修正如下：

def lgCeil(x):  
    if (x <= 1):
        return 0    

    startNum = 0    
    numOfTimes = 0  
    base = 2    
    integer = base * 1  

    while not(integer >= x):    
        startNum = 0    
        numOfTimes = numOfTimes + 1 
        integer = integer * base    

    numOfTimes = numOfTimes + startNum  

    if (integer <= x):  
        base**(numOfTimes+1)    

    return numOfTimes+1 


print (lgCeil(1))   
print (lgCeil(1.1)) 
print (lgCeil(2))   
print (lgCeil(2.1)) 
print (lgCeil(4))   
print (lgCeil(4.1)) 
print (lgCeil(8))   
print (lgCeil(1024))    
print (lgCeil(1025))    
print (lgCeil(2013.15))

虽然我添加了关于x = 1的异常（Klaus D.也提及）（实际上对于0

如果您不打算使用全新版本，这只是总结一下。

Answer 5

这个单行怎么样：

In [7]: len(bin(int(2013.15)).strip("0b"))
Out[7]: 11

Tweak需要支持确切的权力，例如1024，留给读者练习。

LgCeil - 我的教授说我的代码不能用于1

5 个答案: