Compare value of Dataframe column with list value
3
I have a spark dataframe columns 'id' and 'articles' and a list of values 'a_list' as below.
df = spark.createDataFrame([(1, 4), (2, 3), (5, 6)], ("id", "articles"))
a_list = [1, 4, 6]
I am trying to compare list value with value of dataframe column "articles" and if match found updating column 'E' to 1 else 0
I am using "isin" in my code below
df['E'] = df.articles.isin(a_list).astype(int)
Getting
TypeError: unexpected type:
<type 'type'>
What am I missing here ?
python python-3.x pyspark
edited Nov 14 '18 at 0:42
Psidom
123k1284127
asked Nov 14 '18 at 0:40
UmiUmi
626
add a comment |
3
I have a spark dataframe columns 'id' and 'articles' and a list of values 'a_list' as below.
df = spark.createDataFrame([(1, 4), (2, 3), (5, 6)], ("id", "articles"))
a_list = [1, 4, 6]
I am trying to compare list value with value of dataframe column "articles" and if match found updating column 'E' to 1 else 0
I am using "isin" in my code below
df['E'] = df.articles.isin(a_list).astype(int)
Getting
TypeError: unexpected type:
<type 'type'>
What am I missing here ?
python python-3.x pyspark
edited Nov 14 '18 at 0:42
Psidom
123k1284127
asked Nov 14 '18 at 0:40
UmiUmi
626
add a comment |
3
3
3
I have a spark dataframe columns 'id' and 'articles' and a list of values 'a_list' as below.
df = spark.createDataFrame([(1, 4), (2, 3), (5, 6)], ("id", "articles"))
a_list = [1, 4, 6]
I am trying to compare list value with value of dataframe column "articles" and if match found updating column 'E' to 1 else 0
I am using "isin" in my code below
df['E'] = df.articles.isin(a_list).astype(int)
Getting
TypeError: unexpected type:
<type 'type'>
What am I missing here ?
python python-3.x pyspark
edited Nov 14 '18 at 0:42
Psidom
123k1284127
asked Nov 14 '18 at 0:40
UmiUmi
626
I have a spark dataframe columns 'id' and 'articles' and a list of values 'a_list' as below.
df = spark.createDataFrame([(1, 4), (2, 3), (5, 6)], ("id", "articles"))
a_list = [1, 4, 6]
I am trying to compare list value with value of dataframe column "articles" and if match found updating column 'E' to 1 else 0
I am using "isin" in my code below
df['E'] = df.articles.isin(a_list).astype(int)
Getting
TypeError: unexpected type:
<type 'type'>
What am I missing here ?
python python-3.x pyspark
python python-3.x pyspark
edited Nov 14 '18 at 0:42
Psidom
123k1284127
asked Nov 14 '18 at 0:40
UmiUmi
626
edited Nov 14 '18 at 0:42
Psidom
123k1284127
asked Nov 14 '18 at 0:40
UmiUmi
626
edited Nov 14 '18 at 0:42
Psidom
123k1284127
edited Nov 14 '18 at 0:42
Psidom
123k1284127
edited Nov 14 '18 at 0:42
Psidom
123k1284127
123k1284127
asked Nov 14 '18 at 0:40
UmiUmi
626
asked Nov 14 '18 at 0:40
UmiUmi
626
asked Nov 14 '18 at 0:40
UmiUmi
626
626
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
Provide your type as string "int" instead of int which is python's native type that spark doesn't recognize; Also to create a column in spark data frame, use withColumn method instead of direct assignment:
df.withColumn('E', df.articles.isin(a_list).astype('int')).show()
+---+--------+---+
| id|articles| E|
+---+--------+---+
| 1| 4| 1|
| 2| 3| 0|
| 5| 6| 1|
+---+--------+---+
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
1
You can usewhen.otherwiseto conditionally create a column.import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))
– Psidom
Nov 14 '18 at 1:00
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Provide your type as string "int" instead of int which is python's native type that spark doesn't recognize; Also to create a column in spark data frame, use withColumn method instead of direct assignment:
df.withColumn('E', df.articles.isin(a_list).astype('int')).show()
+---+--------+---+
| id|articles| E|
+---+--------+---+
| 1| 4| 1|
| 2| 3| 0|
| 5| 6| 1|
+---+--------+---+
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
1
You can usewhen.otherwiseto conditionally create a column.import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))
– Psidom
Nov 14 '18 at 1:00
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
add a comment |
2
Provide your type as string "int" instead of int which is python's native type that spark doesn't recognize; Also to create a column in spark data frame, use withColumn method instead of direct assignment:
df.withColumn('E', df.articles.isin(a_list).astype('int')).show()
+---+--------+---+
| id|articles| E|
+---+--------+---+
| 1| 4| 1|
| 2| 3| 0|
| 5| 6| 1|
+---+--------+---+
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
1
You can usewhen.otherwiseto conditionally create a column.import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))
– Psidom
Nov 14 '18 at 1:00
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
add a comment |
2
2
2
Provide your type as string "int" instead of int which is python's native type that spark doesn't recognize; Also to create a column in spark data frame, use withColumn method instead of direct assignment:
df.withColumn('E', df.articles.isin(a_list).astype('int')).show()
+---+--------+---+
| id|articles| E|
+---+--------+---+
| 1| 4| 1|
| 2| 3| 0|
| 5| 6| 1|
+---+--------+---+
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
Provide your type as string "int" instead of int which is python's native type that spark doesn't recognize; Also to create a column in spark data frame, use withColumn method instead of direct assignment:
df.withColumn('E', df.articles.isin(a_list).astype('int')).show()
+---+--------+---+
| id|articles| E|
+---+--------+---+
| 1| 4| 1|
| 2| 3| 0|
| 5| 6| 1|
+---+--------+---+
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
answered Nov 14 '18 at 0:46
PsidomPsidom
123k1284127
123k1284127
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
1
You can usewhen.otherwiseto conditionally create a column.import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))
– Psidom
Nov 14 '18 at 1:00
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
add a comment |
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
1
You can usewhen.otherwiseto conditionally create a column.import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))
– Psidom
Nov 14 '18 at 1:00
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
Qq , so instead of updating column with int type , if i were to update it with a text , for 1= Confirmed and for 0 Not confirmed , is there way to do it within above solution ?
– Umi
Nov 14 '18 at 0:53
1
1
You can use
when.otherwise to conditionally create a column. import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))– Psidom
Nov 14 '18 at 1:00
You can use
when.otherwise to conditionally create a column. import pyspark.sql.functions as f; df.withColumn('E', f.when(df.articles.isin(a_list), 'confirmed').otherwise('not confirmed'))– Psidom
Nov 14 '18 at 1:00
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
Sorry to bother again ,if i have a mutiple list like a_list=[4, 10] , b_list=[11,6] , c_list=[3,4] . How can i check against each list and if match found , update column "E" to "Found in a_list" (if found in a_list) or "Found in b_list" (if found in b_list) or "Found in b_list" (if found in c_list)
– Umi
Nov 14 '18 at 1:14
add a comment |
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