Question

我在 pyspark

中有一个如下所示的数据框

df = sqlContext.createDataFrame(
[(1,'Y','Y',0,0,0,2,'Y','N','Y','Y'),
(2,'N','Y',2,1,2,3,'N','Y','Y','N'),
(3,'Y','N',3,1,0,0,'N','N','N','N'),
(4,'N','Y',5,0,1,0,'N','N','N','Y'),
(5,'Y','N',2,2,0,1,'Y','N','N','Y'),
(6,'Y','Y',0,0,3,6,'Y','N','Y','N'),
(7,'N','N',1,1,3,4,'N','Y','N','Y'),
(8,'Y','Y',1,1,2,0,'Y','Y','N','N')
],
('id', 'compatible', 'product', 'ios', 'pc', 'other', 'devices', 'customer', 'subscriber', 'circle', 'smb')
)

现在我想通过连接一些字符串在数据框中创建一个新列 bt_string。我做了如下

import pyspark.sql.functions as f
from datetime import datetime
from time import strftime
from pyspark.sql import Window

# the below values will change as per requirement
job_id = '123'
sess_id = '99'
batch_id = '1'
time_now = datetime.now().strftime('%Y%m%d%H%M%S')

con_string = job_id + sess_id + batch_id + time_now + '000000000000000'

df1 = df.withColumn('bt_string', f.lit(con_string))

现在到数据框，我想为每一行分配一个唯一的编号。我应用了如下所示的 row_number 函数

df2 = df1.withColumn("row_id",f.row_number().over(Window.partitionBy()))

输出如下

df2.show()  

+---+----------+-------+---+---+-----+-------+--------+----------+------+---+--------------------+------+
| id|compatible|product|ios| pc|other|devices|customer|subscriber|circle|smb|           bt_string|row_id|
+---+----------+-------+---+---+-----+-------+--------+----------+------+---+--------------------+------+
|  1|         Y|      Y|  0|  0|    0|      2|       Y|         N|     Y|  Y|12399120210301120...|     1|
|  2|         N|      Y|  2|  1|    2|      3|       N|         Y|     Y|  N|12399120210301120...|     2|
|  3|         Y|      N|  3|  1|    0|      0|       N|         N|     N|  N|12399120210301120...|     3|
|  4|         N|      Y|  5|  0|    1|      0|       N|         N|     N|  Y|12399120210301120...|     4|
|  5|         Y|      N|  2|  2|    0|      1|       Y|         N|     N|  Y|12399120210301120...|     5|
|  6|         Y|      Y|  0|  0|    3|      6|       Y|         N|     Y|  N|12399120210301120...|     6|
|  7|         N|      N|  1|  1|    3|      4|       N|         Y|     N|  Y|12399120210301120...|     7|
|  8|         Y|      Y|  1|  1|    2|      0|       Y|         Y|     N|  N|12399120210301120...|     8|
+---+----------+-------+---+---+-----+-------+--------+----------+------+---+--------------------+------+

现在我想将 row_id 列添加到 bt_string 列。我的意思是像下面

如果 bt_string 行的 1st 是

1239912021030112091500000000000000 then add the corresponding row_id value. 
In the case of first row the value will be 1239912021030112091500000000000001

创建的新列应具有如下值

1239912021030112091500000000000001
1239912021030112091500000000000002
1239912021030112091500000000000003
1239912021030112091500000000000004
1239912021030112091500000000000005
1239912021030112091500000000000006
1239912021030112091500000000000007
1239912021030112091500000000000008

还需要确保列的长度应始终为 35 个字符。

以下字符串无论如何都不应超过 35 个字符的长度

con_string = job_id + sess_id + batch_id + time_now + '000000000000000'

如果超过了 35 个长度的字符，那么我们需要 trim 上面语句中添加的 zeros 的数量。

我怎样才能达到我想要的

Answer 1

按照以下步骤实现您的结果

# import necessary functions
import pyspark.sql.functions as f
from datetime import datetime
from time import strftime
from pyspark.sql import Window

# assign variables as per requirement 
job_id = '123'
sess_id = '99'
batch_id = '1'
time_now = datetime.now().strftime('%Y%m%d%H%M%S')

# Join variables to get desired format of base string
con_string =  job_id + sess_id + batch_id + time_now

# check length of base string and subtract from max length for that column 35 
zero_to_add = 35 - len(con_string)

# Add the numbers of zeros based on the value received above
new_bt_string = con_string + zero_to_add * '0'

# add new column and convert column to decimal and then apply row_number
df1 = df.withColumn('bt_string', f.lit(new_bt_string).cast('decimal(35,0)'))\
    .withColumn("row_id",f.row_number().over(Window.partitionBy()))

# add new column by sum of values from above added columns
df2 = df1.withColumn('bt_id', f.expr('bt_string + row_id'))

将 row_number 添加到数据框 pyspark 中的连接列

1 个答案: