问题描述
是否可以使用聚合框架计算一阶导数?
Is it possible to calculate a first order derivative using the aggregate framework?
例如，我有数据:
{time_series : [10,20,40,70,110]}
我正在尝试获得如下输出:
I'm trying to obtain an output like:
{derivative : [10,20,30,40]}
推荐答案
db.collection.aggregate(
[
{
"$addFields": {
"indexes": {
"$range": [
0,
{
"$size": "$time_series"
}
]
},
"reversedSeries": {
"$reverseArray": "$time_series"
}
}
},
{
"$project": {
"derivatives": {
"$reverseArray": {
"$slice": [
{
"$map": {
"input": {
"$zip": {
"inputs": [
"$reversedSeries",
"$indexes"
]
}
},
"in": {
"$subtract": [
{
"$arrayElemAt": [
"$$this",
0
]
},
{
"$arrayElemAt": [
"$reversedSeries",
{
"$add": [
{
"$arrayElemAt": [
"$$this",
1
]
},
1
]
}
]
}
]
}
}
},
{
"$subtract": [
{
"$size": "$time_series"
},
1
]
}
]
}
},
"time_series": 1
}
}
]
)
我们可以在 3.4+ 版本中使用上述管道来执行此操作.在管道中，我们使用 $addFields 流水线阶段.运算符添加time_series"的元素索引的数组来做文档，我们还反转了时间序列数组并将其添加到文档中分别使用 $range 和 $reverseArray 运算符
We can use the pipeline above in version 3.4+ to do this.
In the pipeline, we use the $addFields pipeline stage. operator to add the array of the "time_series"'s elements index to do document, we also reversed the time series array and add it to the document using respectively the $range and $reverseArray operators
我们在这里反转了数组，因为数组中 p 位置的元素总是大于 p+1 位置的元素，这意味着 [p] - [p+1] <0 并且我们不想使用 $multiply 这里.(请参阅 3.2 版的管道)
We reversed the array here because the element at position p in the array is always greater than the element at position p+1 which means that [p] - [p+1] < 0 and we do not want to use the $multiply here.(see pipeline for version 3.2)
接下来我们用索引数组$zipped 时间序列数据并应用 $map 运算符将 rel="nofollow noreferrer">substract 表达式添加到结果数组.
Next we $zipped the time series data with the indexes array and applied a substract expression to the resulted array using the $map operator.
我们然后$slice将结果从数组中丢弃null/None值并重新反转结果.
We then $slice the result to discard the null/None value from the array and re-reversed the result.
在 3.2 中我们可以使用 $unwind 运算符来展开我们的数组，并通过将文档指定为操作数而不是以 $ 为前缀的传统路径"来包含数组中每个元素的索引.
In 3.2 we can use the $unwind operator to unwind our array and include the index of each element in the array by specifying a document as operand instead of the traditional "path" prefixed by $.
接下来，我们需要 $group 我们的文档并使用 $push 累加器运算符返回一个子文档数组，如下所示:
Next in the pipeline, we need to $group our documents and use the $push accumulator operator to return an array of sub-documents that look like this:
{
"_id" : ObjectId("57c11ddbe860bd0b5df6bc64"),
"time_series" : [
{ "value" : 10, "index" : NumberLong(0) },
{ "value" : 20, "index" : NumberLong(1) },
{ "value" : 40, "index" : NumberLong(2) },
{ "value" : 70, "index" : NumberLong(3) },
{ "value" : 110, "index" : NumberLong(4) }
]
}
<小时>
终于来了 $project 舞台.在这个阶段，我们需要使用 $map 运算符将一系列表达式应用于 $group 阶段中新计算的数组中的每个元素.
Finally comes the $project stage. In this stage, we need to use the $map operator to apply a series of expression to each element in the the newly computed array in the $group stage.
这是 $map 内部发生的事情(将 $map 视为 for 循环)in 表达式:
Here is what is going on inside the $map (see $map as a for loop) in expression:
对于每个子文档，我们使用 value 字段分配给一个变量="nofollow noreferrer">$let 变量运算符.然后我们从数组中下一个元素的value"字段的值中减去它的值.
For each subdocument, we assign the value field to a variable using the $let variable operator. We then subtract it value from the value of the "value" field of the next element in the array.
由于数组中的下一个元素是当前索引处的元素加一，我们所需要的只是 $arrayElemAt 运算符和一个简单的 $addition 当前元素的索引和 1.
Since the next element in the array is the element at the current index plus one, all we need is the help of the $arrayElemAt operator and a simple $addition of the current element's index and 1.
$subtract 表达式返回一个负值，因此我们需要使用 -1"nofollow noreferrer">$multiply 运算符.
The $subtract expression return a negative value so we need to multiply the value by -1 using the $multiply operator.
我们还需要$filter 结果数组，因为它的最后一个元素是 None 或 null.原因是当当前元素是最后一个元素时，$subtract 返回None，因为下一个元素的索引等于数组的大小.
We also need to $filter the resulted array because it the last element is None or null. The reason is that when the current element is the last element, $subtract return None because the index of the next element equal the size of the array.
db.collection.aggregate([
{
"$unwind": {
"path": "$time_series",
"includeArrayIndex": "index"
}
},
{
"$group": {
"_id": "$_id",
"time_series": {
"$push": {
"value": "$time_series",
"index": "$index"
}
}
}
},
{
"$project": {
"time_series": {
"$filter": {
"input": {
"$map": {
"input": "$time_series",
"as": "el",
"in": {
"$multiply": [
{
"$subtract": [
"$$el.value",
{
"$let": {
"vars": {
"nextElement": {
"$arrayElemAt": [
"$time_series",
{
"$add": [
"$$el.index",
1
]
}
]
}
},
"in": "$$nextElement.value"
}
}
]
},
-1
]
}
}
},
"as": "item",
"cond": {
"$gte": [
"$$item",
0
]
}
}
}
}
}
])
<小时>
我认为效率较低的另一个选项是使用 map_reduce 方法.
>>> import pymongo
>>> from bson.code import Code
>>> client = pymongo.MongoClient()
>>> db = client.test
>>> collection = db.collection
>>> mapper = Code("""
... function() {
... var derivatives = [];
... for (var index=1; index<this.time_series.length; index++) {
... derivatives.push(this.time_series[index] - this.time_series[index-1]);
... }
... emit(this._id, derivatives);
... }
... """)
>>> reducer = Code("""
... function(key, value) {}
... """)
>>> for res in collection.map_reduce(mapper, reducer, out={'inline': 1})['results']:
... print(res) # or do something with the document.
...
{'value': [10.0, 20.0, 30.0, 40.0], '_id': ObjectId('57c11ddbe860bd0b5df6bc64')}
<小时>
您还可以检索所有文档并使用 numpy.diff 像这样返回导数:
import numpy as np
for document in collection.find({}, {'time_series': 1}):
result = np.diff(document['time_series'])
这篇关于使用 MongoDB 聚合框架计算一阶导数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！