Question

我在Python 3.6中使用以下代码，该代码返回给定单词在按字母顺序排序的单词的所有可能排列的列表中的位置。例如，请参见下面的测试。提交此代码时，测试环境使用的2.7破坏了我的代码，您知道如何轻松修复它吗？我不熟悉2.7和3.6之间的区别。

我的代码和测试：

def listPosition(word):
    val = [ord(char) - 96 for char in word.lower()]
    minn = 1
    maxx = npermutations(word)
    if all(a >= b for a, b in zip(val, val[1:])):
        return maxx
    if all(a <= b for a, b in zip(val, val[1:])):
        return minn
    for indx in range(len(word)):
            ordi = order(val[indx:],val[indx])
            between = (maxx+1-minn)
            if ordi is 0:
                maxx = maxx - int(between * (1-frequency(val[indx:],val[indx])))
            elif ordi is max([order(val[indx:],i) for i in val[indx:]]):
                minn = minn + int(between * (1-frequency(val[indx:],val[indx])))
            else:
                before = sumfreq(val[indx:],val[indx],'before')
                minn = minn + int(round((between * before),0))
                after = sumfreq(val[indx:],val[indx],'after')
                maxx = maxx - int(round((between * after),0))
    return maxx 
import operator from collections 
import Counter from math 
import factorial from functools 
import reduce 
def npermutations(word):
    num = factorial(len(word))
    mults = Counter(word).values()
    den = reduce(operator.mul, (factorial(v) for v in mults), 1)
    return int(num / den) 
def frequency(val,value):
    f = [val.count(i)/len(val) for i in val]
    indx = val.index(value)
    return f[indx] 
def order(val,value):
    return sorted([i for i in set(val)]).index(value) 
def sumfreq(val,value,BorA):
    if BorA is 'before':
        check = [i for i in set(val) if i < value]
    if BorA is 'after':
        check = [i for i in set(val) if i > value]
    freqs = [frequency(val,i) for i in check]
    return sum(freqs)

tests = ['A','ABAB','AAAB','BAAA','QUESTION','BOOKKEEPER','ABCABC'] 
print(listPosition(tests[0]),"should equal 1") 
print(listPosition(tests[1]),"should equal 2") 
print(listPosition(tests[2]),"should equal 1") 
print(listPosition(tests[3]),"should equal 4") 
print(listPosition(tests[4]),"should equal 24572") 
print(listPosition(tests[5]),"should equal 10743") 
print(listPosition(tests[6]),"should equal 13")

Python 3.6的输出：

1 should equal 1
2 should equal 2
1 should equal 1
4 should equal 4
24572 should equal 24572
10743 should equal 10743
13 should equal 13

提交此代码进行评分时，事实证明测试环境使用的是Python 2.7，这给了我以下输出：

Test Passed
Incorrect list position for: ABAB: 0 should equal 2
Incorrect list position for: QUESTION: 40320 should equal 24572
Incorrect list position for: BOOKKEEPER: 0 should equal 10743
Test Passed
Test Passed

Answer 1

在python 2中，/运算符的行为如下

3 / 2 == 1 # True

在python3中

3 / 2 == 1.5 # True
3 // 2 == 1 # True

编辑

解决此问题的最简单方法是添加

from __future__ import division

在代码的第一行

列出给定单词的所有字符排列的位置，以3.6编写，以2.7分隔

1 个答案: