为什么我的if语句不比较列表项

Question

    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

My code is supposed to increment a sentence's alphabets by 2. ex: a->c, l->n I put the print("hi") statement to check if the if-statement was ever evaluated to be true but it never does. Can someone please tell me why?

Answer 1

is checks object identity . Since you appear to be testing whether two strings have the same value (not are the same object ), you would be better served by == . For example:

if inputlist[x] == alphabet[y]

You can make the same update for your other if statements as well.

Answer 2

The problem is that is compares identity and not equality of strings. Two short strings that are equal may be identical due to some string interning CPython does, but you generally should not build on this behavior. Instead, use == to compare the equality of strings.

Note, that you can do this a lot better using str.translate , with a map created by str.maketrans :

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

You can further use string.ascii_lowercase so you don't need to type the alphabet yourself; or use string.ascii_letters for lower and upper case characters:

>>> table = str.maketrans(string.ascii_letters, string.ascii_letters[2:] + string.ascii_letters[:2])
>>> 'Hello World (This works!)'.translate(table)
'Jgnnq Yqtnf (Vjku yqtmu!)'

Answer 3

Besides of the is thin, you have another problem in your code:

As soon as y == 24 , it will break: First, a will be added and then alphabet[26] - which results in an error.

So change your logic to

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]

This can even improved further:

If you make alphabet = 'abcdefghijklmnopqrstuvwxyz' , you can do

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)]

Answer 4

This might help clear up the "is" keyword. "is" checks the object identity, not the value.

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)))

Output:

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

So, all three have the same value, but only 1 and 3 are identical.