blocks|key|2709567|text|下列措施应能发挥作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2709568|df.groupBy().sum().rdd.map(lambda+x:+x[0]).collect()|code-block|syntax|javascript|2709569|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|$]]

The following should work:

<pre><code>df.groupBy().sum().rdd.map(lambda x: x[0]).collect()
</code></pre>

blocks|key|1075581|text|我认为最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1075582|df.groupBy().sum().collect()|code-block|syntax|javascript|1075583|将返回一个列表。在你的例子中：|1075584|In+[9]:+df.groupBy().sum().collect()[0][0]
Out[9]:+130|1075585|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think the simplest way:

<pre><code>df.groupBy().sum().collect()
</code></pre>

will return a list.
In your example:

<pre><code>In [9]: df.groupBy().sum().collect()[0][0]
Out[9]: 130
</code></pre>

blocks|key|2709629|text|最简单的方法就是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2709630|df.groupBy().sum().collect()|code-block|syntax|javascript|2709631|但是这是非常慢的操作：避免groupByKey，您应该使用RDD和reduceByKey：|offset|length|2709632|df.rdd.map(lambda+x:+(1,x[1])).reduceByKey(lambda+x,y:+x+%2B+y).collect()[0][1]|2709633|我尝试了一个更大的数据集，并测量了处理时间：|2709634|RDD和ReduceByKey：2.23+s|style|BOLD|2709635|GroupByKey:+30.5秒|2709636|entityMap|0|LINK|mutability|MUTABLE|url|https://databricks.gitbooks.io/databricks-spark-knowledge-base/content/best_practices/prefer_reducebykey_over_groupbykey.html^0|0|0|B|C|0|0|0|0|G|6|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@$I|15|J|16|1|17]]|A|$]]|$1|K|3|L|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1A|8|@$I|1B|J|1C|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1D|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

The simplest way really : 

<pre><code>df.groupBy().sum().collect()
</code></pre>

But it is very slow operation: <a href="https://databricks.gitbooks.io/databricks-spark-knowledge-base/content/best_practices/prefer_reducebykey_over_groupbykey.html" rel="noreferrer">Avoid groupByKey</a>, you should use RDD and reduceByKey: 

<pre><code>df.rdd.map(lambda x: (1,x[1])).reduceByKey(lambda x,y: x + y).collect()[0][1]
</code></pre>

I tried on a bigger dataset and i measured the processing time: 

RDD and ReduceByKey : 2.23 s

GroupByKey: 30.5 s

blocks|key|542551|text|这是另一种你能做到的方法。使用agg和collect|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|542552|sum_number+=+df.agg({"Number":"sum"}).collect()[0]

result+=+sum_number["sum(Number)"]|code-block|syntax|javascript|542553|entityMap^0|F|3|J|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

This is another way you can do this. using <code>agg</code> and <code>collect</code>:

<pre><code>sum_number = df.agg({"Number":"sum"}).collect()[0]

result = sum_number["sum(Number)"]
</code></pre>

blocks|key|1075653|text|有时读取csv文件来生成pyspark，也许数字列更改为字符串类型'23'，就像这样，您应该使用pyspark.sql.functions.sum作为int，而不是sum()来获得结果。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1075654|import+pyspark.sql.functions+as+F++++++++++++++++++++++++++++++++++++++++++++++++++++
df.groupBy().agg(F.sum('Number')).show()|code-block|syntax|javascript|1075655|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|$]]

sometimes read a csv file to pyspark Dataframe, maybe the numeric column change to string type '23',like this, you should use pyspark.sql.functions.sum to get the result as int , not sum() 

<pre><code>import pyspark.sql.functions as F 
df.groupBy().agg(F.sum('Number')).show()
</code></pre>

blocks|key|1075671|text|如果您想要一个特定的列：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1075672|import+pyspark.sql.functions+as+F+++++

df.agg(F.sum("my_column")).collect()[0][0]|code-block|syntax|javascript|1075673|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|$]]

If you want a specific column : 

<pre><code>import pyspark.sql.functions as F 

df.agg(F.sum("my_column")).collect()[0][0]
</code></pre>

blocks|key|1075675|text|您也可以尝试使用first()函数。它从dataframe返回第一行，您可以使用索引访问各个列的值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1075676|df.groupBy().sum().first()[0]|code-block|syntax|javascript|1075677|在您的示例中，结果是一个带有单行和列的dataframe，因此上面的代码片段可以工作。|1075678|entityMap^0|8|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can also try using <code>first()</code> function. It returns the first row from the dataframe, and you can access values of respective columns using indices.
<pre><code>df.groupBy().sum().first()[0]
</code></pre>
In your case, the result is a dataframe with single row and column, so above snippet works.

blocks|key|1075745|text|类似于其他答案，但没有使用组或agg。我只需选择所讨论的列，将其和起来，收集它，然后获取前两个索引返回一个int。我之所以选择这样的答案而不是接受的答案，是因为我对这个问题并不熟悉，并且对“Number”列没有在接受的答案中显式地求和感到困惑。如果我不得不在某个时候回来，并试图了解正在发生的事情，下面这样的语法对我来说会更容易理解。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1075746|import+pyspark.sql.functions+as+f+++

df.select(f.sum('Number')).collect()[0][0]|code-block|syntax|javascript|1075747|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|$]]

Similar to other answers, but without the use of a groupby or agg. I just select the column in question, sum it, collect it, and then grab the first two indices to return an int. The only reason I chose this over the accepted answer is I am new to pyspark and was confused that the 'Number' column was not explicitly summed in the accepted answer. If I had to come back after sometime and try to understand what was happening, syntax such as below would be easier for me to follow.
<pre><code>import pyspark.sql.functions as f 

df.select(f.sum('Number')).collect()[0][0]
</code></pre>

blocks|key|1075772|text|选择列作为RDD，使用keys()在行中获取值(或使用.map(lambda+x:+x[0]))，然后使用RDD：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1075773|df.select("Number").rdd.keys().sum()|code-block|syntax|javascript|1075774|使用selectExpr的SQL和|1075775|df.selectExpr("sum(Number)").first()[0]|1075776|entityMap|0|LINK|mutability|MUTABLE|url|https://spark.apache.org/docs/3.1.1/api/python/reference/api/pyspark.sql.DataFrame.selectExpr.html^0|B|6|R|K|0|0|2|A|2|A|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Select column as RDD, abuse <code>keys()</code> to get value in Row (or use <code>.map(lambda x: x[0])</code>), then use RDD sum:
<pre><code>df.select(&quot;Number&quot;).rdd.keys().sum()
</code></pre>
SQL sum using <a href="https://spark.apache.org/docs/3.1.1/api/python/reference/api/pyspark.sql.DataFrame.selectExpr.html" rel="nofollow noreferrer"><code>selectExpr</code></a>:
<pre><code>df.selectExpr(&quot;sum(Number)&quot;).first()[0]
</code></pre>

I have a pyspark dataframe with a column of numbers. I need to sum that column and then have the result return as an int in a python variable. 

<pre><code>df = spark.createDataFrame([("A", 20), ("B", 30), ("D", 80)],["Letter", "Number"])
</code></pre>

I do the following to sum the column. 

<pre><code>df.groupBy().sum()
</code></pre>

But I get a dataframe back. 

<pre><code>+-----------+
|sum(Number)|
+-----------+
| 130|
+-----------+
</code></pre>

I would 130 returned as an int stored in a variable to be used else where in the program. 

<pre><code>result = 130
</code></pre>

PySpark - Sum a column in dataframe and return results as int

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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我有一列数字的电火花数据。我需要对该列进行求和，然后在python变量中将结果作为int返回。df = spark.createDataFrame([("A", 20), ("B", 30), ("D", 80)],["Letter", "Number"])我做下面的总结这一栏。df.groupBy().sum()但我...

问PySpark -对数据格式中的列进行求和，并以int形式返回结果
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问PySpark -对数据格式中的列进行求和，并以int形式返回结果EN