Question

我有两个数据帧，df1和df2：

df1.show()
+---+--------+-----+----+--------+
|cA |   cB   |  cC | cD |   cE   |
+---+--------+-----+----+--------+
|  A|   abc  | 0.1 | 0.0|   0    |
|  B|   def  | 0.15| 0.5|   0    |
|  C|   ghi  | 0.2 | 0.2|   1    |
|  D|   jkl  | 1.1 | 0.1|   0    |
|  E|   mno  | 0.1 | 0.1|   0    |
+---+--------+-----+----+--------+


df2.show()
+---+--------+-----+----+--------+
|cA |   cB   |  cH | cI |   cJ   |
+---+--------+-----+----+--------+
|  A|   abc  | a   | b  |   ?    |
|  C|   ghi  | a   | c  |   ?    |
+---+--------+-----+----+--------+

如果在cE中引用了该行，我想更新df1中的1列并将其设置为df2。每个记录由cA和cB列标识。

下面是所需的输出；请注意，第一条记录的cE值已更新为1：

+---+--------+-----+----+--------+
|cA |   cB   |  cC | cD |   cE   |
+---+--------+-----+----+--------+
|  A|   abc  | 0.1 | 0.0|   1    |
|  B|   def  | 0.15| 0.5|   0    |
|  C|   ghi  | 0.2 | 0.2|   1    |
|  D|   jkl  | 1.1 | 0.1|   0    |
|  E|   mno  | 0.1 | 0.1|   0    |
+---+--------+-----+----+--------+

Answer 1

这是我的答案。

这是scala代码-抱歉-我没有安装python。希望有帮助。

import org.apache.spark.sql._
import org.apache.spark.sql.functions._

val ss = SparkSession.builder().master("local").getOrCreate()

import ss.implicits._

val seq1 = Seq(
  ("A", "abc", 0.1, 0.0, 0),
  ("B", "def", 0.15, 0.5, 0),
  ("C", "ghi", 0.2, 0.2, 1),
  ("D", "jkl", 1.1, 0.1, 0),
  ("E", "mno", 0.1, 0.1, 0)
)

val seq2 = Seq(
  ("A", "abc", "a", "b", "?"),
  ("C", "ghi", "a", "c", "?")
)


val df1 = ss.sparkContext.makeRDD(seq1).toDF("cA", "cB", "cC", "cD", "cE")
val df2 = ss.sparkContext.makeRDD(seq2).toDF("cA", "cB", "cH", "cI", "cJ")


val joined = df1.join(df2, (df1("cA") === df2("cA")).and(df1("cB") === df2("cB")), "left")

val res = joined.withColumn("newCe",
  when(df2("cA").isNull.and(joined("cE") === lit(0)), lit(0)).otherwise(lit(1)))


res.select(df1("cA"), df1("cB"), df1("cC"), df1("cD"), res("newCe"))
  .withColumnRenamed("newCe", "cE")
  .show

对我来说，输出是：

+---+---+----+---+---+
| cA| cB|  cC| cD| cE|
+---+---+----+---+---+
|  E|mno| 0.1|0.1|  0|
|  B|def|0.15|0.5|  0|
|  C|ghi| 0.2|0.2|  1|
|  A|abc| 0.1|0.0|  1|
|  D|jkl| 1.1|0.1|  0|
+---+---+----+---+---+

Answer 2

如果存在基于另一列更新列值的情况，那么when子句会派上用场。请参阅when and else条款。

import pyspark.sql.functions as F
df3=df1.join(df2,(df1.cA==df2.cA)&(df1.cB==df2.cB),"full").withColumn('cE',F.when((df1.cA==df2.cA)&(df1.cB==df2.cB),1).otherwise(0)).select(df1.cA,df1.cB,df1.cC,df1.cD,'cE')
df3.show()
+---+---+----+---+---+
| cA| cB|  cC| cD| cE|
+---+---+----+---+---+
|  E|mno| 0.1|0.1|  0|
|  B|def|0.15|0.5|  0|
|  C|ghi| 0.2|0.2|  1|
|  A|abc| 0.1|0.0|  1|
|  D|jkl| 1.1|0.1|  0|
+---+---+----+---+---+

Answer 3

使用加入，您可以做自己想做的事情：

df1 = pd.DataFrame({ 'cA' : ['A', 'B', 'C', 'D', 'E'], 'cB' : ['abc', 'def', 'ghi', 'jkl', 'mno'], 'cE' : [0,0,1, 0, 0]})
df2 = pd.DataFrame({ 'cA' : ['A', 'C'], 'cB' : ['abc', 'ghi'], 'cE' : ['?','?']})

# join
df = df1.join(df2.set_index(['cA', 'cB']),  lsuffix='_df1', rsuffix='_df2', on=['cA', 'cB'])

# nan values indicates rows that are not present in both dataframes
df.loc[~df['cE_df2'].isna(), 'cE_df2'] = 1
df.loc[df['cE_df2'].isna(), 'cE_df2'] = 0

df1['cE'] = df['cE_df2']

输出：

    cA  cB  cE
0   A   abc 1
1   B   def 0
2   C   ghi 1
3   D   jkl 0
4   E   mno 0

Answer 4

尝试

for i in df2.values:
    df1.loc[(df1.cA==i[0]) & (df1.cB == i[1]),['cE']] = 1

根据另一个数据框中的值更新数据框中的列

4 个答案: