比较pyspark中的两个数据集 [英] Compare two datasets in pyspark

问题描述

我有2个数据集.

示例数据集1:

id     |   model |   first_name   |      last_name
-----------------------------------------------------------
1234   |   32    |    456765      |   [456700,987565]
-----------------------------------------------------------
4539   |   20    |    123211      |   [893456,123456]
-----------------------------------------------------------

有时first_name和last_name列之一为空.

Some times one of the columns first_name and last_name is empty.

示例数据集2:

number  |  matricule   | name       |    model
----------------------------------------------------------
AA      |  0009        |  456765    |     32
----------------------------------------------------------
AA      |  0009        |  893456    |     32
----------------------------------------------------------
AA      |  0009        |  456700    |     32
----------------------------------------------------------
AA      |  0008        |  456700    |     32
----------------------------------------------------------
AA      |  0008        |  987565    |     32

对于一个matricule，我们可以找到更多的name和model，就像上面的示例一样. 我应该怎么做:

For one matricule we can find more name and model, like in my example just above. What I should do:

对于数据集1中的每一行，我提取3列:模型，first_name和last_name，并在数据集2中查找它们(如果存在/根据矩阵元素匹配).

For each row from the Dataset 1, I take the 3 columns: model, first_name and last_name and look for them in Dataset 2 if exist / match according the matricule elements.

我应该比较:

• 按模型划分的模型==>如果模型(数据集1)存在于模型(数据集2)==>匹配中

• model by model ==> if model (dataset 1) exist in model (dataset 2) ==> match

如果名字中存在名字==>则不匹配.如果名字中不存在first_name ==>匹配

if first_name exist in name ==> no match. If first_name not exist in name ==> match

如果在name ==>匹配项中存在last_name.当我有两个last_name值时，这两个值都应存在于要匹配的数据集2的名称中.

if last_name exist in name ==> match. When I have two values of last_name, the both should exist in name of dataset 2 to be matched.

示例:

数据集1中的行1为:

id     |   model |   first_name   |      last_name
------------------------------------------------------
1234   |   32    |    456765      |   [456700,987565]

对于数据集2中的矩阵0009，我有:

For matricule 0009 in dataset 2, I have:

number  |  matricule   | name       |    model
----------------------------------------------------------
AA      |  0009        |  456765    |     32
----------------------------------------------------------
AA      |  0009        |  893456    |     32
----------------------------------------------------------
AA      |  0009        |  456700    |     32

所以:

first_name(456765)

first_name (456765) is exist in name of dataset 2 when matriule =0009 ==> no match

姓氏，仅存在456700 ==>不匹配

last_name, only 456700 is exist ==> no match

模型(32)==>匹配

model (32) is exist in model of dataset 2 ==> match

因此，我跳过了矩阵0009.然后将数据集1中的第二行与矩阵0008的元素进行比较.

So I skip the matricule 0009. And pass to compare second line in dataset 1 with the elements of matricule 0008.

对于数据集2中的矩阵0008，我有:

For matricule 0008 in dataset 2, I have:

----------------------------------------------------------
AA      |  0008        |  456700    |     32
----------------------------------------------------------
AA      |  0008        |  987565    |     32

我们总是在数据集1的第一行中

Always we are in the first rows of dataset 1:

first_name(456765)

first_name (456765) is not exist in name of dataset 2 when matricule=0008 ==> match

last_name，当矩阵= 0008，==>匹配时，两个值都存在于数据集2的名称中

last_name, the both values are exist in name of dataset 2 when matricule = 0008, ==> match

模型

当我找到所有匹配项时，我将创建一个包含以下内容的新数据集:

When I find all match, I create a new dataset contain:

number | id     |  matricule
-----------------------------------
AA     | 1234   | 0008
-----------------------------------

我希望我很清楚.有人可以帮我.

I hope that I was clear. Someone can help me please.

推荐答案

您可以在符合条件的情况下使用join.

You can use join on the conditions of matching.

首先，您可以将第二个DataFrame分组，并将name列收集到一个列表中:

First, you can group by the second DataFrame and collect name column into a list:

df2 = df2.groupBy("number", "model", "matricule").agg(collect_list("name").alias("names"))
f2.show(truncate=False)

#+------+-----+---------+------------------------+
#|number|model|matricule|names                   |
#+------+-----+---------+------------------------+
#|AA    |32   |0009     |[456765, 893456, 456700]|
#|AA    |32   |0008     |[456700, 987565]        |
#+------+-----+---------+------------------------+

现在，加入df1和df2.对于条件1和2，检查起来有点简单. 对于第三个，您可以使用 array_except 可从Spark 2.4+开始使用(last_name列中不应有names中没有的元素，反之亦然):

Now, join df1 and df2. For conditions 1 and 2, it is somehow simple to check. For the third one, you can use array_except avaliable from Spark 2.4+ (there should be no elements from last_name column that are not in names and vice-versa):

join_condition = (col("df1.model") == col("df2.model")) \
                 & ~expr("array_contains(df2.names, df1.first_name)") \
                 & (size(expr("array_except(df2.names, df1.last_name)")) == lit(0)) \
                 & (size(expr("array_except(df1.last_name, df2.names)")) == lit(0))


df_result = df1.alias("df1").join(df2.alias("df2"), join_condition)

最后，从联接结果中选择所需的列:

Finally, select desired columns from the join result:

df_result.select("number", "id", "matricule").show(truncate=False)

#+------+----+---------+
#|number|id  |matricule|
#+------+----+---------+
#|AA    |1234|0008     |
#+------+----+---------+

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