• • 巨蟒：**

我有一个 Dataframe ，我正在应用一个lambda函数来检查基于列值的条件。

In Pandas it looks like this(Example):

new_df = df1.merge(df2, how='left', left_on='lkey', right_on='rkey')

  lkey value_x rkey value_y col1 col2 col3 col4 col5 
0  foo     one  foo    five  0    1     3    0   5    
1  foo     one  foo     NaN  1    0     2    4   0    
2  bar     two  bar     six  2    6     3    0   0    
3  foo    five  foo    five  7    2     0    0   0    
4  foo    five  foo     NaN  2    0     0    0   0   
5  bbb    four  bar    two   0    0     0    0   0      

def get_final_au(row):
    if row['col5'] == 0: 
        if row['col4'] == 0: 
            if row['col3'] == 0: 
                if row['col2'] == 0: 
                    return 'NOT FOUND'
                else:
                    return row['col2']
            else:
                return row['col3']
        else:
            return row['col4']
    else:
         return row['col5']

new_df['col6'] = new_df.apply (lambda row: get_final_au(row),axis=1)

Expected Output:

  lkey value_x rkey value_y col1 col2 col3 col4 col5 col6
0  foo     one  foo    five  0    1     3    0   5    5
1  foo     one  foo     NaN  1    0     2    4   0    4
2  bar     two  bar     six  2    6     3    0   0    3
3  foo    five  foo    five  7    2     0    0   0    2
4  foo    five  foo     NaN  2    0     0    0   0   Not FOUND
5  bbb    four  bar    two   0    0     0    0   0   Not FOUND

• • 游乐园：**

在Pyspark中，我将如何做类似的事情？

我已经尝试过这个，但得到错误。请建议

from pyspark.sql.functions import udf
def get_final_au(row):
    if row['col5'] != 0:
        return row['col5']
    elif row['col4'] != 0:
        return row['col4']
    elif row['col3'] != 0:
        return row['col3']
    elif row['col2'] != 0:
        return row['col2']
    else:
        return 'NOT FOUND'
UDF_NAME = udf(lambda row: get_final_au(row), StringType())
new_df.withColumn('col6', UDF_NAME('col5','col4','col3','col2')).show(2,False)