我正在使用 python 和 Tornado 学习 mongodb。我有一个 mongodb 集合，当我这样做的时候

db.cal.find()

{     
    "Pid" : "5652f92761be0b14889d9854",
    "Registration" : "TN 56 HD 6766",
    "Vid" : "56543ed261be0b0a60a896c9",
    "Period" : "10-2015",
    "AOs": [
        "14-10-2015",
        "15-10-2015",
        "18-10-2015",
        "14-10-2015",
        "15-10-2015",
        "18-10-2015"
    ],
    "Booked": [
        "5-10-2015",
        "7-10-2015",
        "8-10-2015",
        "5-10-2015",
        "7-10-2015",
        "8-10-2015"
    ],
    "NA": [
        "1-10-2015",
        "2-10-2015",
        "3-10-2015",
        "4-10-2015",
        "1-10-2015",
        "2-10-2015",
        "3-10-2015",
        "4-10-2015"
    ],

    "AOr": [
        "23-10-2015",
        "27-10-2015",
        "23-10-2015",
        "27-10-2015"
    ]
}

我需要一个操作来删除 Booked,NA,AOs,AOr 中的重复值。最后应该是

{
     "Pid" : "5652f92761be0b14889d9854",
      "Registration" : "TN 56 HD 6766",
      "Vid" : "56543ed261be0b0a60a896c9",
      "AOs": [
        "14-10-2015",
        "15-10-2015",
        "18-10-2015",

      ],
      "Booked": [
        "5-10-2015",
        "7-10-2015",
        "8-10-2015",

      ],

      "NA": [
        "1-10-2015",
        "2-10-2015",
        "3-10-2015",
        "4-10-2015",

      ],

      "AOr": [
        "23-10-2015",
        "27-10-2015",

      ]
}

我如何在 mongodb 中实现这一点？

最佳答案

工作解决方案

我已经创建了一个基于 JavaScript 的工作解决方案，它在 mongo shell 上可用:

var codes = ["AOs", "Booked", "NA", "AOr"]

// Use bulk operations for efficiency
var bulk = db.dupes.initializeUnorderedBulkOp()

db.dupes.find().forEach(
  function(doc) {

    // Needed to prevent unnecessary operatations
    changed = false
    codes.forEach(
      function(code) {
        var values = doc[code]
        var uniq = []

        for (var i = 0; i < values.length; i++) {
          // If the current value can not be found, it is unique
          // in the "uniq" array after insertion
          if (uniq.indexOf(values[i]) == -1 ){
            uniq.push(values[i])
          }
        }

        doc[code] = uniq

        if (uniq.length < values.length) {
          changed = true
        }

      }
    )

    // Update the document only if something was changed
    if (changed) {
      bulk.find({"_id":doc._id}).updateOne(doc)
    }
  }
)

// Apply all changes
bulk.execute()

包含示例输入的结果文档:

replset:PRIMARY> db.dupes.find().pretty()
{
  "_id" : ObjectId("567931aefefcd72d0523777b"),
  "Pid" : "5652f92761be0b14889d9854",
  "Registration" : "TN 56 HD 6766",
  "Vid" : "56543ed261be0b0a60a896c9",
  "Period" : "10-2015",
  "AOs" : [
    "14-10-2015",
    "15-10-2015",
    "18-10-2015"
  ],
  "Booked" : [
    "5-10-2015",
    "7-10-2015",
    "8-10-2015"
  ],
  "NA" : [
    "1-10-2015",
    "2-10-2015",
    "3-10-2015",
    "4-10-2015"
  ],
  "AOr" : [
    "23-10-2015",
    "27-10-2015"
  ]
}

通过 dropDups 使用索引

这根本行不通。首先，根据 3.0 版，此选项不再存在。既然我们已经发布了 3.2，我们应该找到一种可移植的方式。

其次，即使有 dropDups，文档也明确指出:

dropDups boolean : MongoDB indexes only the first occurrence of a key and removes all documents from the collection that contain subsequent occurrences of that key.

因此，如果有另一个文档在其中一个帐单代码中具有与前一个相同的值，则整个文档将被删除。

关于python - 删除mongodb中的重复值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34414422/