blocks|key|4077239|text|以下map创建由customer聚合的price的映射|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4077240|def+vals+=+(0..(-1%2Bmap.customer.size()))
+++++++++++.collect{['name':map.customer[it],+'price':+map.price[it]]}
+++++++++++.groupBy{it['name']}
+++++++++++.collectEntries{[(it.key):+it.value.collect{it['price']}.sum()]}|code-block|syntax|javascript|4077241|这将导致：|4077242|[Clinton:63000.0,+Mark:89000.0,+Antony:56000.0]|4077243|它本质上是一个迭代，使用从0到map.customer.size()+-+1的一系列数字，然后是具有值总和的group-by。|4077244|entityMap^0|8|8|J|5|0|0|0|0|D|1|F|N|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

The followint creates a map of <code>price</code> aggregated by <code>customer</code>:

<pre><code>def vals = (0..(-1+map.customer.size()))
 .collect{['name':map.customer[it], 'price': map.price[it]]}
 .groupBy{it['name']}
 .collectEntries{[(it.key): it.value.collect{it['price']}.sum()]}
</code></pre>

That results in:

<pre><code>[Clinton:63000.0, Mark:89000.0, Antony:56000.0]
</code></pre>

It's essentially an iteration using a range of numbers from <code>0</code> to <code>map.customer.size() - 1</code>, followed by a group-by with sum of values.

blocks|key|4077265|text|在这样的问题中，我们应该始终计划将每个条目作为列表中的单独对象，以使其在列表中直观且将来易于操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4077266|在这种情况下，可以自然地获得相同的结果|4077267|def+list+=+[

++++++++[customer:+'Clinton',+price:+15000.0],
++++++++[customer:+'Clinton',+price:+27000.0],
++++++++[customer:+'Mark',++++price:+28000.0],
++++++++[customer:+'Antony',++price:+56000.0],
++++++++[customer:+'Clinton',+price:+21000.0],
++++++++[customer:+'Mark',++++price:+61000.0]
]


def+map+=+list.groupBy({it.customer}).collectEntries+{k,+v+->+[k,+v.price.sum()]}

map.each+{println+it}|code-block|syntax|javascript|4077268|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In the problems like this, we should always plan having every entry as separate object inside a list to make it intuitive and future easy manipulation in the list.

In that case the same result can be obtained in naturally

<pre><code>def list = [

 [customer: 'Clinton', price: 15000.0],
 [customer: 'Clinton', price: 27000.0],
 [customer: 'Mark', price: 28000.0],
 [customer: 'Antony', price: 56000.0],
 [customer: 'Clinton', price: 21000.0],
 [customer: 'Mark', price: 61000.0]
]


def map = list.groupBy({it.customer}).collectEntries {k, v -&gt; [k, v.price.sum()]}

map.each {println it}
</code></pre>

blocks|key|4075982|text|您可以从两个列表上的转置开始，以获得customer和price的元组。从那里它基本上与其他答案一样(按客户分组，与客户建立地图，并汇总价格)。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4075983|def+data+=+[
++++customer:['Clinton',+'Clinton',+'Mark',+'Antony',+'Clinton',+'Mark'],+
++++price:[15000.0,+27000.0,+28000.0,+56000.0,+21000.0,+61000.0]
]

println(
++++[data.customer,+data.price].transpose().groupBy{+c,+p+->+c+}.collectEntries{+c,+ps+->+[c,+ps*.last().sum()+]+}
)
//+=>+[Clinton:63000.0,+Mark:89000.0,+Antony:56000.0]|code-block|syntax|javascript|4075984|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can start with a transpose on both lists to get tuples of customer and price. From there its basically like the other answers (group by customer, build map with customer and summed up prices). E.g.:

<pre><code>def data = [
 customer:['Clinton', 'Clinton', 'Mark', 'Antony', 'Clinton', 'Mark'], 
 price:[15000.0, 27000.0, 28000.0, 56000.0, 21000.0, 61000.0]
]

println(
 [data.customer, data.price].transpose().groupBy{ c, p -&gt; c }.collectEntries{ c, ps -&gt; [c, ps*.last().sum() ] }
)
// =&gt; [Clinton:63000.0, Mark:89000.0, Antony:56000.0]
</code></pre>

blocks|key|4077371|text|这个版本是从@cfrick的答案派生出来的，并有一个求和的替代方案。使用带有默认值的映射并不是“功能性”的/declarative，但我认为代码更容易理解(即维护)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4077372|给定：|4077373|def+map+=+[
++++customer:+['Clinton',+'Clinton',+'Mark',+'Antony',+'Clinton',+'Mark'],
++++price:+[15000.0,+27000.0,+28000.0,+56000.0,+21000.0,+61000.0]
]|code-block|syntax|javascript|4077374|方法：|4077375|def+getSumMap+=+{+data+->
++++def+result+=+[:].withDefault+{+c+->+0.0+}

++++[data.customer,+data.price].transpose().each+{+c,+p+->+result[c]+%2B=+p+}

++++result
}

assert+63000.0+==+getSumMap(map)['Clinton']
assert+89000.0+==+getSumMap(map)['Mark']
assert+56000.0+==+getSumMap(map)['Antony']|4077376|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This version is derived from @cfrick's answer, with an alternative to the summation. Using a map with default values isn't as "functional"/declarative, but IMHO the code is arguably easier to grok later (i.e. maintenance):

Given:

<pre><code>def map = [
 customer: ['Clinton', 'Clinton', 'Mark', 'Antony', 'Clinton', 'Mark'],
 price: [15000.0, 27000.0, 28000.0, 56000.0, 21000.0, 61000.0]
]
</code></pre>

Approach:

<pre><code>def getSumMap = { data -&gt;
 def result = [:].withDefault { c -&gt; 0.0 }

 [data.customer, data.price].transpose().each { c, p -&gt; result[c] += p }

 result
}

assert 63000.0 == getSumMap(map)['Clinton']
assert 89000.0 == getSumMap(map)['Mark']
assert 56000.0 == getSumMap(map)['Antony']
</code></pre>

I have a map with two keys (<code>customer</code> and <code>price</code>) like the one shown below:

<pre><code>[
 customer: ['Clinton', 'Clinton', 'Mark', 'Antony', 'Clinton', 'Mark'], 
 price: [15000.0, 27000.0, 28000.0, 56000.0, 21000.0, 61000.0]
]
</code></pre>

<code>customer</code> and <code>price</code> values are mapped by their index positione i.e first name from <code>customer</code> list maps with the first price and so on.

Example

<pre><code>Cliton price is 15000.0
Cliton price 27000.0
Mark price 28000.0
Antony price 56000.0
Clinton price 21000.0
Mark price 61000.0
</code></pre>

I would like to sum up price grouped by names. Expected output:

<pre><code>Clinton price 63000
Mark price 89000
Antony price 56000
</code></pre>

Are there any built-in functions to achieve this in Groovy, or do I need to iterate over the map and sum values by writing my own functions?

How to sum map values based on two keys?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个带有两个键(customer和price)的映射，如下所示：[    customer: ['Clinton', 'Clinton', 'Mark', 'Antony', 'Clinton', 'Mark'],     price: [15000.0, 27000.0, 28000.0, 56000.0, 21...

问如何根据两个键对映射值求和？
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问如何根据两个键对映射值求和？
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