Python 如何避免'；1'；当传入为“无”时

我想要的结果是，当incoming2和incoming3为无时，应该避免在incoming1和incoming4之间打印两个“1”

def string_con(self,combined):
    if combined is none:
       return '1'
    else:
       return str(combined)

if incoming['col1'] or incoming['col1'] or incoming['col3'] or incoming['col4']:    
combined = self.string_con(incoming['col1'])+'1'+self.string_con(incoming['col2'])+'1'+self.string_con(incoming['col3'])+'1'+self.string_con(incoming['col4'])

---

你能试一下吗

req_keys = ['col1', 'col2', 'col3', 'col4']
all_list = [incoming[i] for i in req_keys]
all_list = [i for i in all_list if i]
print('1'.join(all_list))

例如：

incoming = {}
incoming['col1'] = 'a'
incoming['col2'] = None
incoming['col3'] = 'c'
incoming['col4'] = None

输出：

a1c

a1d

另一个例子：

incoming = {}
incoming['col1'] = 'a'
incoming['col2'] = None
incoming['col3'] = None
incoming['col4'] = 'd'

输出：

a1c

a1d

---

如果传入['col1']或传入['col1']，您的代码中有一个输入错误-->

。。。猜第二个应该是
incoming['col2']
。。。这是你的错误的根源吗？有人能把英语讲干净吗？我真的不确定问的是什么。
None
，而不是
None
。资本化很重要。