Python 如何在字典中使用reduce
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how to use reduce with dictionary
提问by Salvador Dali
I have some problem understanding how to use reduce with dictionaries in python. For example I have the following dictionary.
我在理解如何在 python 中使用 reduce 和字典时遇到了一些问题。例如，我有以下字典。
{1: 3, 2: 1, 3: 2}
{1: 3, 2: 1, 3: 2}
and I am trying to calculate the following:
我正在尝试计算以下内容：
s = 0
for i in h:
s += h[i] * (h[i] - 1)
This works as expected (I get: 8), but I my attempt to convert it to reduce paradigm fails:
reduce(lambda x, y: x + y * (y - 1), h), but I am getting the wrong answer.
这按预期工作（我得到：）8，但我尝试将其转换为减少范式失败：
reduce(lambda x, y: x + y * (y - 1), h)，但我得到了错误的答案。
I assume this is because I am using keys, not values. How can I convert my code to reduce properly?
我认为这是因为我使用的是键，而不是值。如何转换我的代码以正确减少？
采纳答案by Abhijit
You need to iterate over the dictionary while reducing it with an initial value of zero.
您需要迭代字典，同时将其初始值为零。
Note, iterating over a dictionary, actually iterates over the keys so you need to index the dictionary to get the value
注意，遍历字典，实际上是遍历键，所以你需要索引字典来获取值
reduce(lambda x, key:x + h[key] * (h[key] - 1), h, 0)
Alternatively, as you are only interested in the values of the dictionary, caring least about the key, just iterate on the values of the dictionary
或者，由于您只对字典的值感兴趣，而不关心键，只需迭代字典的值
Python 2.X
蟒蛇2.X
reduce(lambda x, value:x + value * (value - 1), h.itervalues(), 0)
Python 3.X
蟒蛇 3.X
reduce(lambda x, value:x + value * (value - 1), h.values(), 0)
