python - 代码将整数读取为 float

Question

我的代码将 iMid 读取为 float 并给出 TypeError，即使将其包装在整数函数中也是如此。另外，是否还有另一种方法来查找中间值的索引，这比我在这里尝试的方法更容易？

def isIn(char, aStr):
'''
char: a single character
aStr: an alphabetized string

returns: True if char is in aStr; False otherwise
'''
# Your code here
import numpy as np
def iMid(x):
    '''
    x : a string

    returns: index of the middle value of the string

    '''

    if len(x) % 2 == 0:
        return int(np.mean(len(x)/2, (len(x)+2)/2)) #wrapped the 
                                                    # answer for iMid 
                                                    #in the integer function
    else:
        return int((len(x)+1)/2)

if char == aStr[iMid] or char == aStr: #iMid is not being interpreted as an integer
    return True
elif char < aStr[iMid]:
    return isIn(char, aStr[0:aStr[iMid]]) 
else:
    return isIn(char, aStr[aStr[iMid]:])

print(isIn('c', "abcd"))

Answer 1

在

if char == aStr[iMid] or char == aStr: #iMid is not being interpreted as an integer

iMid 不是整数。这就是函数。

您需要调用该函数来获取它返回的整数。

if char == aStr[iMid(aStr)] or char == aStr: #iMid is called and returns an integer