替换Spark dataframe中列名的特殊字符

Question

我的输入spark-dataframe名为df，

+---------------+----------------+-----------------------+
|Main_CustomerID|126+ Concentrate|2.5 Ethylhexyl_Acrylate|
+---------------+----------------+-----------------------+
|         725153|             3.0|                    2.0|
|         873008|             4.0|                    1.0|
|         625109|             1.0|                    0.0|
+---------------+----------------+-----------------------+

我需要从df的列名中删除特殊字符，如下所示，

• 删除+
• 将空间替换为underscore
• 将dot替换为underscore

所以我的df应该是

+---------------+---------------+-----------------------+
|Main_CustomerID|126_Concentrate|2_5_Ethylhexyl_Acrylate|
+---------------+---------------+-----------------------+
|         725153|            3.0|                    2.0|
|         873008|            4.0|                    1.0|
|         625109|            1.0|                    0.0|
+---------------+---------------+-----------------------+

我用Scala实现了这一点

var tableWithColumnsRenamed = df

for (field <- tableWithColumnsRenamed.columns) {
      tableWithColumnsRenamed = tableWithColumnsRenamed
        .withColumnRenamed(field, field.replaceAll("\\.", "_"))
    }
for (field <- tableWithColumnsRenamed.columns) {
      tableWithColumnsRenamed = tableWithColumnsRenamed
        .withColumnRenamed(field, field.replaceAll("\\+", ""))
    }
for (field <- tableWithColumnsRenamed.columns) {
      tableWithColumnsRenamed = tableWithColumnsRenamed
        .withColumnRenamed(field, field.replaceAll(" ", "_"))
    }

df = tableWithColumnsRenamed

当我用，

for (field <- tableWithColumnsRenamed.columns) {
      tableWithColumnsRenamed = tableWithColumnsRenamed
        .withColumnRenamed(field, field.replaceAll("\\.", "_"))
    .withColumnRenamed(field, field.replaceAll("\\+", ""))
    .withColumnRenamed(field, field.replaceAll(" ", "_"))
    }

我得到的第一个列名为126 Concentrate，而不是126_Concentrate

但是我不喜欢这个替换的3 for循环。我能找到解决办法吗？

Answer 1

df
  .columns
  .foldLeft(df){(newdf, colname) =>
    newdf.withColumnRenamed(colname, colname.replace(" ", "_").replace(".", "_"))
  }
  .show

Answer 2

您可以使用withColumnRenamed regex replaceAllIn和foldLeft，如下所示

val columns = df.columns

val regex = """[+._, ]+"""
val replacingColumns = columns.map(regex.r.replaceAllIn(_, "_"))

val resultDF = replacingColumns.zip(columns).foldLeft(df){(tempdf, name) => tempdf.withColumnRenamed(name._2, name._1)}

resultDF.show(false)

这应该会给你

+---------------+---------------+-----------------------+
|Main_CustomerID|126_Concentrate|2_5_Ethylhexyl_Acrylate|
+---------------+---------------+-----------------------+
|725153         |3.0            |2.0                    |
|873008         |4.0            |1.0                    |
|625109         |1.0            |0.0                    |
+---------------+---------------+-----------------------+

我希望答案是有帮助的。

Answer 3

在java中，您可以使用df.columns()迭代列名，并用string replaceAll(regexPattern, IntendedCharreplacement)替换每个头字符串。

然后使用withColumnRenamed(headerName, correctedHeaderName)重命名df头。

伊格-

for (String headerName : dataset.columns()) {
    String correctedHeaderName = headerName.replaceAll(" ","_").replaceAll("+","_");
    dataset = dataset.withColumnRenamed(headerName, correctedHeaderName);
}
dataset.show();