MongoDB的数据转换

我的测试集MongoDB的数据转换

{ "_id" : 0, "Animals" : "cat", "activity" : "sleep" } 
{ "_id" : 1, "Animals" : "dog", "activity" : "run" } 
{ "_id" : 2, "Animals" : "cow", "activity" : "play" } 
{ "_id" : 3, "Animals" : "cow", "activity" : "sleep" } 
{ "_id" : 4, "Animals" : "cow", "activity" : "run" } 
{ "_id" : 5, "Animals" : "dog", "activity" : "play" } 
{ "_id" : 6, "Animals" : "cat", "activity" : "run" }

动物后发现动物和独特的价值活动

db.test.distinct("Animals") 
[ "cat", "dog", "cow" ] 
db.test.distinct("service") 
[ "sleep", "run", "play" ]

==猫 - > {猫：1，狗：0，牛：0}同其他

我想改变这种格式

{ "_id" : 0, "cat" : 1, "dog" : 0, "cow" : 0, "sleep" : 1, "run" : 0, "play": 0 } 
{ "_id" : 1, "cat" : 0, "dog" : 1, "cow" : 0, "sleep" : 0, "run" : 1, "play": 0 } 
{ "_id" : 2, "cat" : 0, "dog" : 0, "cow" : 1, "sleep" : 0, "run" : 0, "play": 1 } 
{ "_id" : 3, "cat" : 0, "dog" : 0, "cow" : 1, "sleep" : 1, "run" : 0, "play": 0 } 
{ "_id" : 4, "cat" : 0, "dog" : 0, "cow" : 1, "sleep" : 0, "run" : 1, "play": 0 } 
{ "_id" : 5, "cat" : 0, "dog" : 1, "cow" : 0, "sleep" : 0, "run" : 0, "play": 1 } 
{ "_id" : 6, "cat" : 1, "dog" : 0, "cow" : 0, "sleep" : 0, "run" : 1, "play": 0 }

我该怎么做？

来源

2015-07-20 mohamedzajith

与聚合框架如下我会做到这一点：

组由_id
算上动物
算上活动
排序结果按升序排列

db.test.aggregate([ { 
$group: { 
_id: "$_id", 
cat: {$sum: {$cond: [{"$eq": ["$Animals", "cat"]}, 1, 0 ] } }, 
dog: {$sum: {$cond: [{"$eq": ["$Animals", "dog"]}, 1, 0 ] } }, 
cow: {$sum: {$cond: [{"$eq": ["$Animals", "cow"]}, 1, 0 ] } }, 
sleep: {$sum: {$cond: [{"$eq": ["$activity", "sleep"]}, 1, 0 ] } }, 
run: {$sum: {$cond: [{"$eq": ["$activity", "run"]}, 1, 0 ] } }, 
play: {$sum: {$cond: [{"$eq": ["$activity", "play"]}, 1, 0 ] } } 
} }, 
{ $sort: {_id: 1}} 
])

在蒙戈外壳结果看起来像您正在寻找的东西：

{ "_id" : 0, "cat" : 1, "dog" : 0, "cow" : 0, "sleep" : 1, "run" : 0, "play" : 0 } 
{ "_id" : 1, "cat" : 0, "dog" : 1, "cow" : 0, "sleep" : 0, "run" : 1, "play" : 0 } 
{ "_id" : 2, "cat" : 0, "dog" : 0, "cow" : 1, "sleep" : 0, "run" : 0, "play" : 1 } 
{ "_id" : 3, "cat" : 0, "dog" : 0, "cow" : 1, "sleep" : 1, "run" : 0, "play" : 0 } 
{ "_id" : 4, "cat" : 0, "dog" : 0, "cow" : 1, "sleep" : 0, "run" : 1, "play" : 0 } 
{ "_id" : 5, "cat" : 0, "dog" : 1, "cow" : 0, "sleep" : 0, "run" : 0, "play" : 1 } 
{ "_id" : 6, "cat" : 1, "dog" : 0, "cow" : 0, "sleep" : 0, "run" : 1, "play" : 0 }

来源

2015-07-20 13:30:29 Elyasin

Elyasin我想代码帮助我吗？ – mohamedzajith

我想避免为你编写代码。这很耗时，并且不确定您的学习体验是否更有效率。我建议你尝试编码自己，然后当你被困时帮助你，是吗？ – Elyasin

基本上他告诉你要做的是从现有表中提取数据并以编程方式插入到新表中。我也同意这种做法，这是一次性格式转换。 –

我想告诉你用aggregation如下：

db.test.aggregate([ 
    {$group : { 
     _id : $_id, 
     cat : {$cond : {if: { $eq: [ "$Animals", "cat" ] }, then: 1, else: 0}}, 
     dog : {$cond : {if: { $eq: [ "$Animals", "dog" ] }, then: 1, else: 0}}, 
     cow : {$cond : {if: { $eq: [ "$Animals", "cow" ] }, then: 1, else: 0}}, 
     sleep : {$cond : {if: { $eq: [ "$activity", "sleep" ] }, then: 1, else: 0}}, 
     run : {$cond : {if: { $eq: [ "$activity", "run" ] }, then: 1, else: 0}}, 
     play : {$cond : {if: { $eq: [ "$activity", "play" ] }, then: 1, else: 0}} 
    }} 
]);

这个片段中使用$cond集结算子+ $eq比较运算符。

来源

2015-07-20 13:58:18 Ramin

MongoDB的数据转换

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