Question

我有一个字典定义如下：

SN = {}
SN[0] = {'verb' : 1 }

在我的函数中，我按如下方式生成3 print：

print graph
print state
print graph[state]

（我对这些变量不做任何其他事情）并返回：

SN  
0  
S

这怎么可能？为什么不返回

SN
0
{'verb' : 1}

整个代码：

abstraction= {}
abstraction['de'] = ['déterminant']
abstraction['le'] = ['déterminant']
abstraction['un'] = ['déterminant']
abstraction['beau'] = ['adjectif', 'substantif']
abstraction['dodu'] = ['adjectif', 'substantif']
abstraction['grand'] = ['adjectif', 'substantif']
abstraction['méchant'] = ['adjectif', 'substantif']
abstraction['noirs'] = ['adjectif', 'substantif']
abstraction['petit'] = ['adjectif', 'substantif']
abstraction['desseins'] = ['substantif']
abstraction['loup'] = ['substantif']
abstraction['poulet'] = ['substantif']
abstraction['gosse'] = ["n'importe quoi"]
abstraction['mange'] = ['verbe']
abstraction['dort'] = ['verbe']


SN = {}
SN[0] = {'déterminant' : 1 }
SN[1] = {'adjectif' : 1, 'substantif' : 2 }
SN[2] = {'adjectif' : 3, '' : 'ok' }
SN[3] = {'' : 'ok' }

SV = {}
SV[0] = {'verbe' : 1}
SV[1] = {'transitions' : 'SN', '' : 'ok'}
SV[2] = {'' : 'ok'}


def transitions(data, graph, state = 0, transit = []) :
    print 'data avt if data :'
    print data
    if data : 
        if 'transitions' in graph[state] :
            return transitions(data, graph[state]['transitions'], 0, transit)
        symbol = data[0]
        if symbol not in abstraction.keys() : 
            return ['impossible, un des mots n\'est pas reconnu par l\'automate'] 
        for a in abstraction[symbol] : # loop on abstractions
            print graph
            print state
            print graph[state]
            if a in graph[state] :
                state = graph[state][a]
                return transitions(data[1:], graph, state, transit + [a])
        else :  
            return transit + [symbol] + ['impossible'] 
    else : 
        if '' in graph[state] :
            return transit + [graph[state]['']]
        else : return transit + ['impossible']

Answer 1

我认为您的问题是graph == "SN"，而不是（正如您所期望的那样）graph == SN。

换句话说，graph引用的str对象的值为"SN"，不 dict对象也被引用名称SN。

因此graph[0]是字符串"SN"中的第一个字符，即字母"S"。

在graph == SN的情况下，来自

的输出

print graph
print state
print graph[state]

将是：

{0: {'verb': 1}} # note: not 'SN'
0
{'verb': 1}

修改：

现在您已发布了代码，此部分：

SN = {} SN[0] = {'déterminant' : 1 } SN[1] = {'adjectif' : 1, 'substantif' : 2 } SN[2] = {'adjectif' : 3, '' : 'ok' } SN[3] = {'' : 'ok' }

创建字典

SN = {0: {'déterminant': 1}, 1: {'adjectif': 1, 'substantif': 2}, 2: {'adjectif': 3, '': 'ok'}, 3: {'': 'ok'}}

然而，在下一部分：

SV[1] = {'transitions' : 'SN', '' : 'ok'}

您将字符串'SN' 分配给密钥'transitions'，而不是实际字典SN。应该是：

SV[1] = {'transitions': SN, '': 'ok'} # ^ no quote marks

此外，由于您的所有键都是从零开始的整数，您可以考虑使用列表而不是字典，例如：

SN = [] SN.append({'déterminant': 1}) SN.append({'adjectif': 1, 'substantif': 2}) ...

以下三个打印时难以理解的错误

1 个答案: