Question

我有字典，其值作为列表：

    myDict =  {'id1': ["name1","Type1","Value_1"],
    'id2': ["name2","Type1","Value_2"],
    'id3': ["name1","Type2","Value_3"],
    'id4': ["name1","Type1","Value_4"]
    }

我想遍历字典，看看“名称和类型”对是否已在列表中-将“ Type 1”值替换为其他任何值，得到的字典将是：

    myDict =  {'id1': ["name1","Type1","Value_1"],
    'id2': ["name2","Type1","Value_2"],
    'id3': ["name1","Type2","Value_3"],
    'id4': ["name1","Modified_Type 1","Value_4"]
    }

目前不知道如何使用Python处理它

主要是关于比较value [0]和value [1]以及在其他列表中是否都相同的问题-替换它。

我试图遍历现有字典并比较其值是否不在newDictionary中，但是显然我正在检查这些值是否单独存在于newDict值中，而不是成对存在：

    myDict =  {'id1': ["name1","Type1","Value_1"],
    'id2': ["name2","Type1","Value_2"],
    'id3': ["name1","Type2","Value_3"],
    'id4': ["name1","Type1","Value_4"]
    }
    newDict = {}
    for key, value in myDict.items():
        if value[0] not in newDict.values() and value[1] not in newDict.values():
    newDict[key] = value
    else:
        newDict[key] = [value[0],"Some modified value",value[2]]
  print (newDict)

Answer 1

目前尚不清楚您想要什么，因为您的结果两次包含Type1 ...但这是一种开始正确路径的方法。

听起来您想对ID进行排序。这样您就可以获得像这样的键的排序列表：

keys = sorted(myDict) #thanks @abarnert

然后遍历并检查类型：

existingTypes = []
for key in keys:
    theType = myDict[key][1]
    if theType in existingTypes:
        myDict[key][1] = "Modified_" + theType
    else:
        existingTypes.push(theType)

编辑-更新您更新的问题：

这也许不是最干净的，但它可以工作：

myDict =  {'id1': ["name1","Type1","Value_1"],
'id2': ["name2","Type1","Value_2"],
'id3': ["name1","Type2","Value_3"],
'id4': ["name1","Type1","Value_4"]
}

newDict = {}
for key in sorted(myDict):
    value = myDict[key]
    valuesExist = False

    for newValue in newDict.values():
        if value[0] == newValue[0]  and value[1] == newValue[1]:
            valuesExist = True

    if not valuesExist:
        newDict[key] = value
    else:
        newDict[key] = [value[0],"Some modified value",value[2]]

print (newDict)

检查列表中是否有重复的对，并替换其中一个值

1 个答案: