未使用Python 3.6解决加密问题

Question

我一直在研究密码学问题。我试图用OrderedDict编写的第一个代码是这样的。你知道吗

from collections import OrderedDict

alphabet = OrderedDict(
    [(u'a', u'ç'), (u'b', u'd'), (u'c', u'e'), (u'ç', u'f'), (u'd', u'g'), (u'e', u'g'), (u'f', u'h'),
     (u'g', u'i'), (u'g', u'i'), (u'h', u'j'), (u'i', u'k'), (u'i', u'l'), (u'j', u'm'), (u'k', u'n'),
     (u'l', u'o'), (u'm', u'ö'), (u'n', u'p'), (u'o', u'r'), (u'ö', u's'), (u'p', u'ş'), (u'r', u't'),
     (u's', u'u'), (u'ş', u'ü'), (u't', u'v'), (u'u', u'y'), (u'ü', u'z'), (u'v', u'a'),
     (u'y', u'b'), (u'z',u'c'), (9, '')])


text = 'öğtjçdç9dğp9grnvrt9jyuğbkp9içokoğr9pyp9nçtkbğt9iypoğtk9gycğpoğgkikpk9gybgyö9ağ9uğpkp9kekp9drboğ9dkt9urty9\
            jçcktoçgkö9içokoğr9gç9uvçmç9dçuoçöçn9kekp9vğn9bçşöçp9iğtğnğp9\
                çpçjvçtk9dyogyiyp9ağ9dy9öğvkp9grubçukpk9rnybçdkogkikp9\
                    çoirtkvöçoçtk9kuvğgkikp9gkogğ9nrgoçöçp9ağ9dçuoçtkpç9çpçjvçt9\
                        ağ9ngvççmnçç9bçcçtçn9nrgoçtkpk9uktnğvkp9kphr9çgğtğukpğ9öçko9çvöçpgkt9\jçgk9irtğbkö9uğpk'

out = []

for k, v in alphabet.items():
    for i in list(text):
        if i == v:
            out.append(alphabet[v])

outlast = (''.join(out))

print(outlast)

我无法比较“文本”和字母值的列表。我想“文本的每个元素（I）比较值，如果它是'I==values'将I附加到'out'列表，但作为键！”我在用PyCharm 2017.2.2你能帮我吗？你知道吗

实际上密码是3。”“土耳其字母表上升了3倍”

这是错误的输出：

ffffffffffffffffffffffffffffffffffffffffffffgggggggggggiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiijllllllllllmmmmmmööpppppppppppprrrrrrrrrrrrrrrrrrsssssssssssşşşşşşşşşşşşşşşşşşşşşşşşşşşştttttttttttttüvvvvvvvvvvvvvvvvvvvvyyyyyyyyyyyyaaaaaaaaaaabbbbbbbbbbbbççççdddddddddeee

预期产量：

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