_corrupt_record error when reading a JSON file into Spark

Question

I've got this JSON file

{
    "a": 1, 
    "b": 2
}

which has been obtained with Python json.dump method. Now, I want to read this file into a DataFrame in Spark, using pyspark. Following documentation, I'm doing this

sc = SparkContext()

sqlc = SQLContext(sc)

df = sqlc.read.json('my_file.json')

print df.show()

The print statement spits out this though:

+---------------+
|_corrupt_record|
+---------------+
|              {|
|       "a": 1, |
|         "b": 2|
|              }|
+---------------+

Anyone knows what's going on and why it is not interpreting the file correctly?

Solution 1

You need to have one json object per row in your input file, see http://spark.apache.org/docs/latest/api/python/pyspark.sql.html#pyspark.sql.DataFrameReader.json

If your json file looks like this it will give you the expected dataframe:

{ "a": 1, "b": 2 }
{ "a": 3, "b": 4 }

....
df.show()
+---+---+
|  a|  b|
+---+---+
|  1|  2|
|  3|  4|
+---+---+

Solution 2

If you want to leave your JSON file as it is (without stripping new lines characters \n), include multiLine=True keyword argument

sc = SparkContext() 
sqlc = SQLContext(sc)

df = sqlc.read.json('my_file.json', multiLine=True)

print df.show()

Solution 3

In Spark 2.2+ you can read json file of multiline using following command.

val dataframe = spark.read.option("multiline",true).json( " filePath ")

if there is json object per line then,

val dataframe = spark.read.json(filepath)

Solution 4

Adding to @Bernhard's great answer

# original file was written with pretty-print inside a list
with open("pretty-printed.json") as jsonfile:
    js = json.load(jsonfile)      

# write a new file with one object per line
with open("flattened.json", 'a') as outfile:
    for d in js:
        json.dump(d, outfile)
        outfile.write('\n')

View more solutions

Comments

[M.Rez]

How can I fix it if my JSON file is huge (a couple of 100K rows) and it has a lot of new lines in between the records (columns or features)? thanks.

[Ankita Mehta]

This is scala, not python.

[ttimasdf]

Maybe use jq to reformat(compact) the file? @M.Rez

[Skippy le Grand Gourou]

@ttimasdf Namely, using jq -c option.

[Christian Singer]

Still works in Python however spark.read.option("multiline",True).json('filePath')