0 
    alphabet =["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
def decoder(input):
    inputlist = list(input)
    inputlength = len(input)
    alphabetlength = len(alphabet)
    result = "Decoded Sentence: "
    for x in range(inputlength):
        for y in range(alphabetlength):
            if inputlist[x] is alphabet[y]:
                print ("hi")
                if y == 24:
                    result += "a"
                if y == 25:
                    result += "b"
                else:
                    result += alphabet[y+2]
            if inputlist[x] is "(":
                result += "("
            if inputlist[x] is ")":
                result += ")"
            if inputlist[x] is ".":
                result += "."
            if inputlist[x] is " ":
                result += " "
    return result

我的代码应该将句子的字母增加 2。例如：a->c, l->n 我使用 print("hi") 语句来检查 if 语句是否被评估为真，但它从来没有. 有人可以告诉我为什么吗？

4

4 回答 4

5

is检查对象身份。由于您似乎正在测试两个字符串是否具有相同的值（不是相同的object），因此您最好使用==. 例如：

if inputlist[x] == alphabet[y]

您也可以对其他if语句进行相同的更新。

于 2013-11-04T20:38:25.717 回答
4

问题是is比较字符串的同一性而不是相等性。由于 CPython 的某些字符串实习，两个相等的短字符串可能相同，但您通常不应该建立在这种行为上。相反，用于==比较字符串的相等性。

str.translate请注意，您可以使用创建的地图更好地做到这一点str.maketrans：

>>> table = str.maketrans('abcdefghijklmopqrstuvwxyz', 'cdefghijklmopqrstuvwxyzab')
>>> 'hello world'.translate(table)
'jgooq yqtof'

您可以进一步使用string.ascii_lowercase，因此您无需自己输入字母；或string.ascii_letters用于小写和大写字符：

>>> table = str.maketrans(string.ascii_letters, string.ascii_letters[2:] + string.ascii_letters[:2])
>>> 'Hello World (This works!)'.translate(table)
'Jgnnq Yqtnf (Vjku yqtmu!)'
于 2013-11-04T20:41:12.587 回答
1

除了is瘦之外，您的代码中还有另一个问题：

一旦y == 24，它将中断：首先，a将被添加，然后alphabet[26]- 导致错误。

所以改变你的逻辑

for inp in inputlist:
    if inp in "(). ":
        result += inp
    else: # very important
        for y in range(alphabetlength):
            if inp == alphabet[y]:
                if y == 24:
                    result += "a"
                elif y == 25: # elif instead of if!
                    result += "b"
                else:
                    result += alphabet[y+2]

这甚至可以进一步改进：

如果你做alphabet = 'abcdefghijklmnopqrstuvwxyz'，你可以做

for inp in inputlist:
    if inp in "(). ":
        result += inp
    else: # very important
        idx = alphabet.find(inp)
        if idx >= 0: # found
            result += alphabet[(idx + 2) % len(alphabet)]
于 2013-11-04T20:49:53.560 回答
0

这可能有助于清除“is”关键字。“is”检查对象身份，而不是值。

var_1 = ('a', 'b')
var_2 = ('a', 'b') # same content, but different object
var_3 = var_1 # same object
n = lambda v: ' ' if v else ' not '
print('{0} has{2}the same value as {1}'.format('var_1', 'var_2', n(var_1 == var_2)))
print('{0} has{2}the same value as {1}'.format('var_2', 'var_3', n(var_2 == var_3)))
print('{0} has{2}the same value as {1}'.format('var_3', 'var_1', n(var_3 == var_1)))
print('{0} is{2}the same object as {1}'.format('var_1', 'var_2', n(var_1 is var_2)))
print('{0} is{2}the same object as {1}'.format('var_2', 'var_3', n(var_2 is var_3)))
print('{0} is{2}the same object as {1}'.format('var_3', 'var_1', n(var_3 is var_1)))

输出：

var_1 has the same value as var_2
var_2 has the same value as var_3
var_3 has the same value as var_1
var_1 is not the same object as var_2
var_2 is not the same object as var_3
var_3 is the same object as var_1

因此，所有三个都具有相同的值，但只有 1 和 3 相同。

于 2013-11-04T20:45:14.597 回答