admin管理员组

文章数量:1123344

In pyspark I am able to get the filename in a column using:

df = spark.read.option("delimiter", ";").load(inlees_pad, format='csv', header=True)
df = df.withColumn("filename", input_file_name())

I try the same using sparklyr in R:

sigma_raw <- spark_read_csv(
  sc,
  name = "comma_decimal_df",
  path = inlees_pad,
  delimiter = ";",            # Use ";" as the delimiter
  header = TRUE               # Include headers if the file has them
) %>% mutate(filename = input_file_name())

However in sparklyr the column filename remains empty without an error. Does anyone know how to fix this?

本文标签： rHow to get original filename in column using sparklyr sparkreadcsvStack Overflow

版权声明：本文标题：r - How to get original filename in column using sparklyr spark_read_csv? - Stack Overflow 内容由网友自发贡献，该文观点仅代表作者本人， 转载请联系作者并注明出处：http://www.betaflare.com/web/1736562622a1944663.html， 本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌抄袭侵权/违法违规的内容，一经查实，本站将立刻删除。