我有两个从csv加载的Spark数据框:
mapping_fields(带有映射名称的df):
new_name old_name
A aa
B bb
C cc
和
aa bb cc dd
1 2 3 43
12 21 4 37
转变为:
A B C D
1 2 3
12 21 4
由于dd在原始表中没有任何映射,因此D列应该具有所有空值。
如何在不将mapping_df转换为字典并单独检查映射名称的情况下如何做到这一点? (这意味着我必须收集mapping_fields并检查,这与我分布式处理所有数据集的用例相矛盾)
谢谢!
答案 0 :(得分:1)
从here借用melt
你可以:
from pyspark.sql import functions as f
mapping_fields = spark.createDataFrame(
[("A", "aa"), ("B", "bb"), ("C", "cc")],
("new_name", "old_name"))
df = spark.createDataFrame(
[(1, 2, 3, 43), (12, 21, 4, 37)],
("aa", "bb", "cc", "dd"))
(melt(df.withColumn("id", f.monotonically_increasing_id()),
id_vars=["id"], value_vars=df.columns, var_name="old_name")
.join(mapping_fields, ["old_name"], "left_outer")
.withColumn("value", f.when(f.col("new_name").isNotNull(), col("value")))
.withColumn("new_name", f.coalesce("new_name", f.upper(col("old_name"))))
.groupBy("id")
.pivot("new_name")
.agg(f.first("value"))
.drop("id")
.show())
+---+---+---+----+
| A| B| C| DD|
+---+---+---+----+
| 1| 2| 3|null|
| 12| 21| 4|null|
+---+---+---+----+
但在你的描述中没有任何理由这样做。因为列数相当有限,我宁愿:
mapping = dict(
mapping_fields
.filter(f.col("old_name").isin(df.columns))
.select("old_name", "new_name").collect())
df.select([
(f.lit(None).cast(t) if c not in mapping else col(c)).alias(mapping.get(c, c.upper()))
for (c, t) in df.dtypes])
+---+---+---+----+
| A| B| C| DD|
+---+---+---+----+
| 1| 2| 3|null|
| 12| 21| 4|null|
+---+---+---+----+
在一天结束时,您应该在提供性能或可伸缩性改进时使用分布式处理。在这里它会做相反的事情并使你的代码过于复杂。
忽略不匹配:
(melt(df.withColumn("id", f.monotonically_increasing_id()),
id_vars=["id"], value_vars=df.columns, var_name="old_name")
.join(mapping_fields, ["old_name"])
.groupBy("id")
.pivot("new_name")
.agg(f.first("value"))
.drop("id")
.show())
或
df.select([
col(c).alias(mapping.get(c))
for (c, t) in df.dtypes if c in mapping])
答案 1 :(得分:0)
我尝试了一个简单的for循环,希望这也有帮助。
from pyspark.sql import functions as F
l1 = [('A','aa'),('B','bb'),('C','cc')]
l2 = [(1,2,3,43),(12,21,4,37)]
df1 = spark.createDataFrame(l1,['new_name','old_name'])
df2 = spark.createDataFrame(l2,['aa','bb','cc','dd'])
print df1.show()
+--------+--------+
|new_name|old_name|
+--------+--------+
| A| aa|
| B| bb|
| C| cc|
+--------+--------+
>>> df2.show()
+---+---+---+---+
| aa| bb| cc| dd|
+---+---+---+---+
| 1| 2| 3| 43|
| 12| 21| 4| 37|
+---+---+---+---+
当您需要具有空值的缺失列时,
>>>cols = df2.columns
>>> for i in cols:
val = df1.where(df1['old_name'] == i).first()
if val is not None:
df2 = df2.withColumnRenamed(i,val['new_name'])
else:
df2 = df2.withColumn(i,F.lit(None))
>>> df2.show()
+---+---+---+----+
| A| B| C| dd|
+---+---+---+----+
| 1| 2| 3|null|
| 12| 21| 4|null|
+---+---+---+----+
当我们只需要映射列时,更改else部分,
else:
df2 = df2.drop(i)
>>> df2.show()
+---+---+---+
| A| B| C|
+---+---+---+
| 1| 2| 3|
| 12| 21| 4|
+---+---+---+
这会改变原来的df2数据帧。