无法匹配python中字典中的键

Question

我把下面这句话指定为

dict1 = {('Q6', 'pic'): 'yes'}

我正在尝试匹配dict1中的键Q6,pic。我使用以下代码

dict1Index = 6
key = 'Q' + str(dict1Index)
dictPicKey = key + "," + "pic"
if dictPicKey in dict1 : 

print "*****"

它与输出不匹配

Answer 1

您只是在创建一个带有逗号的字符串。你要找的钥匙是薄纱。试试这个：

dict1Index = 6 
key = 'Q' + str(dict1Index)
dictPicKey = (key, "pic") 
if dictPicKey in dict1 : 
    print "*****"