Polars 从虚拟变量转换回

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英文:

Polars Convert Back From Dummies

问题

# 在pandas中,我可以使用[`from_dummies`](https://pandas.pydata.org/docs/reference/api/pandas.from_dummies.html)方法来反转独热编码。在polars中似乎没有内置的方法来做到这一点。这里是一个基本的例子:

pl.DataFrame({
  "col1_hi": [0,0,0,1,1],
  "col1_med": [0,0,1,0,0],
  "col1_lo": [1,1,0,0,0],
  "col2_yes": [1,1,0,1,0],
  "col2_no": [0,0,1,0,1],
})

反转to_dummies操作应该得到类似这样的结果:

pl.DataFrame({
  "col1": ["lo", "lo", "med", "hi", "hi"],
  "col2": ["yes", "yes", "no", "yes", "no"],
})

我的第一反应是使用pivot。我该如何实现这个功能?

英文:

In pandas I can use the from_dummies method to reverse one-hot encoding. There doesn't seem to be a built in method for this in polars. Here is a basic example:

pl.DataFrame({
  "col1_hi": [0,0,0,1,1],
  "col1_med": [0,0,1,0,0],
  "col1_lo": [1,1,0,0,0],
  "col2_yes": [1,1,0,1,0],
  "col2_no": [0,0,1,0,1],
})

┌─────────┬──────────┬─────────┬──────────┬─────────┐
│ col1_hi ┆ col1_med ┆ col1_lo ┆ col2_yes ┆ col2_no │
│ ---     ┆ ---      ┆ ---     ┆ ---      ┆ ---     │
│ i64     ┆ i64      ┆ i64     ┆ i64      ┆ i64     │
╞═════════╪══════════╪═════════╪══════════╪═════════╡
│ 0       ┆ 0        ┆ 1       ┆ 1        ┆ 0       │
│ 0       ┆ 0        ┆ 1       ┆ 1        ┆ 0       │
│ 0       ┆ 1        ┆ 0       ┆ 0        ┆ 1       │
│ 1       ┆ 0        ┆ 0       ┆ 1        ┆ 0       │
│ 1       ┆ 0        ┆ 0       ┆ 0        ┆ 1       │
└─────────┴──────────┴─────────┴──────────┴─────────┘

Reversing the to_dummies operation should result in something like this:

┌──────┬──────┐
│ col1 ┆ col2 │
│ ---  ┆ ---  │
│ str  ┆ str  │
╞══════╪══════╡
│ lo   ┆ yes  │
│ lo   ┆ yes  │
│ med  ┆ no   │
│ hi   ┆ yes  │
│ hi   ┆ no   │
└──────┴──────┘

My first thought was to use a pivot. How could I go about implementing this functionality?

答案1

得分: 4

你可以利用 pl.coalesce

(df
 .with_columns(
    pl.when(pl.col(col) == 1)
      .then(pl.lit(col).str.extract(r"([^_]+$)"))
      .alias(col) 
    for col in df.columns)
 .select(
    pl.coalesce(pl.col(f"^{prefix}_.+$")).alias(prefix) 
    for prefix in dict.fromkeys(
       col.rsplit("_", maxsplit=1)[0]
       for col in df.columns
    )
))
形状:(5,2)
┌──────┬──────┐
│ col1 ┆ col2 │
│ ---  ┆ ---  │
│ str  ┆ str  │
╞══════╪══════╡
│ lo   ┆ yes  │
│ lo   ┆ yes  │
│ med  ┆ no   │
│ hi   ┆ yes  │
│ hi   ┆ no   │
└──────┴──────┘

更新: @Rodalm's方法 更简洁:

def from_dummies(df, separator="_"):
    col_exprs = {}
    
    for col in df.columns:
        name, value = col.rsplit(separator, maxsplit=1)
        expr = pl.when(pl.col(col) == 1).then(value) 
        col_exprs.setdefault(name, []).append(expr)

    return df.select(
        pl.coalesce(exprs) # 保留每行的第一个非空表达式值
          .alias(name)
        for name, exprs in col_exprs.items()
    )
英文:

You could utilize pl.coalesce

(df
 .with_columns(
    pl.when(pl.col(col) == 1)
      .then(pl.lit(col).str.extract(r"([^_]+$)"))
      .alias(col) 
    for col in df.columns)
 .select(
    pl.coalesce(pl.col(f"^{prefix}_.+$")).alias(prefix) 
    for prefix in dict.fromkeys(
       col.rsplit("_", maxsplit=1)[0]
       for col in df.columns
    )
))
shape: (5, 2)
┌──────┬──────┐
│ col1 ┆ col2 │
│ ---  ┆ ---  │
│ str  ┆ str  │
╞══════╪══════╡
│ lo   ┆ yes  │
│ lo   ┆ yes  │
│ med  ┆ no   │
│ hi   ┆ yes  │
│ hi   ┆ no   │
└──────┴──────┘

Update: @Rodalm's approach is much neater:

def from_dummies(df, separator="_"):
    col_exprs = {}
    
    for col in df.columns:
        name, value = col.rsplit(separator, maxsplit=1)
        expr = pl.when(pl.col(col) == 1).then(value) 
        col_exprs.setdefault(name, []).append(expr)

    return df.select(
        pl.coalesce(exprs) # keep the first non-null expression value by row
          .alias(name)
        for name, exprs in col_exprs.items()
    )

答案2

得分: 2

使用 pl.coalesce 的方法,类似于 @jqurious's answer

from collections import defaultdict
import polars as pl

df = pl.DataFrame({
  "col1_hi": [0,0,0,1,1],
  "col1_med": [0,0,1,0,0],
  "col1_lo": [1,1,0,0,0],
  "col2_yes": [1,1,0,1,0],
  "col2_no": [0,0,1,0,1],
})

def from_dummies(df, sep="_"):

    col_exprs = defaultdict(list)
    for col in df.columns:
        name, value = col.split(sep)
        expr = pl.when(pl.col(col) == 1).then(value) # null otherwise
        col_exprs[name].append(expr)

    res = df.select(**{
        name: pl.coalesce(exprs) # keep the first non-null expression value by row
        for name, exprs in col_exprs.items()
    })
    return res

或者是泛化 @warwick12's approach,使用多个 whenthen 连接的方法:

def from_dummies(df, sep="_"):

    col_exprs = {}
    for col in df.columns:
        name, value = col.split(sep)
        if name not in col_exprs:
            col_exprs[name] = pl.when(pl.col(col) == 1).then(value)
        else:
            col_exprs[name] = col_exprs[name].when(pl.col(col) == 1).then(value)
 
    return df.select(**col_exprs)

输出:

>>> from_dummies(df)

shape: (5, 2)
┌──────┬──────┐
 col1  col2 
 ---   ---  
 str   str  
╞══════╪══════╡
 lo    yes  
 lo    yes  
 med   no   
 hi    yes  
 hi    no   
└──────┴──────┘
英文:

A similar approach to @jqurious's answer using pl.coalesce:

from collections import defaultdict
import polars as pl

df = pl.DataFrame({
  "col1_hi": [0,0,0,1,1],
  "col1_med": [0,0,1,0,0],
  "col1_lo": [1,1,0,0,0],
  "col2_yes": [1,1,0,1,0],
  "col2_no": [0,0,1,0,1],
})

def from_dummies(df, sep="_"):

    col_exprs = defaultdict(list)
    for col in df.columns:
        name, value = col.split(sep)
        expr = pl.when(pl.col(col) == 1).then(value) # null otherwise
        col_exprs[name].append(expr)

    res = df.select(**{
        name: pl.coalesce(exprs) # keep the first non-null expression value by row
        for name, exprs in col_exprs.items()
    })
    return res

Or generalizing @warwick12's approach using multiple when and thens chained:

def from_dummies(df, sep="_"):

    col_exprs = {}
    for col in df.columns:
        name, value = col.split(sep)
        if name not in col_exprs:
            col_exprs[name] = pl.when(pl.col(col) == 1).then(value)
        else:
            col_exprs[name] = col_exprs[name].when(pl.col(col) == 1).then(value)
 
    return df.select(**col_exprs)

Output:

>>> from_dummies(df)

shape: (5, 2)
┌──────┬──────┐
 col1  col2 
 ---   ---  
 str   str  
╞══════╪══════╡
 lo    yes  
 lo    yes  
 med   no   
 hi    yes  
 hi    no   
└──────┴──────┘

答案3

得分: 1

你可以使用 pl.when()、pl.col() 和 pl.lit() 方法将包含虚拟变量的 Polars DataFrame 转换回原始格式。这将每列的虚拟变量映射回其原始值。

# 创建虚拟变量的 DataFrame
df = pl.DataFrame({
    "col1_hi": [0,0,0,1,1],
    "col1_med": [0,0,1,0,0],
    "col1_lo": [1,1,0,0,0],
    "col2_yes": [1,1,0,1,0],
    "col2_no": [0,0,1,0,1],
})

# 将虚拟变量映射回原始值
df = df.select([
    pl.when(pl.col("col1_hi") == 1).then(pl.lit("hi"))
    .when(pl.col("col1_med") == 1).then(pl.lit("med")).otherwise("lo").alias("col1"),
    pl.when(pl.col("col2_yes") == 1).then(pl.lit("yes")).otherwise("no").alias("col2")
])

# 显示原始 DataFrame
print(df)
英文:

You can use the pl.when(), pl.col() and pl.lit() methods to convert a polars DataFrame with dummy variables back to the original format. This map's each column's dummies back to their original values.

import polars as pl

# Create dummy variable DataFrame
df = pl.DataFrame({
    "col1_hi": [0,0,0,1,1],
    "col1_med": [0,0,1,0,0],
    "col1_lo": [1,1,0,0,0],
    "col2_yes": [1,1,0,1,0],
    "col2_no": [0,0,1,0,1],
})

# Map dummies back to original values
df = df.select([
    pl.when(pl.col("col1_hi") == 1).then(pl.lit("hi"))
    .when(pl.col("col1_med") == 1).then(pl.lit("med")).otherwise("lo").alias("col1"),
    pl.when(pl.col("col2_yes") == 1).then(pl.lit("yes")).otherwise("no").alias("col2")
])

# Display original DataFrame
print(df)

Output:

shape: (5, 2)
┌──────┬──────┐
│ col1 ┆ col2 │
│ ---  ┆ ---  │
│ str  ┆ str  │
╞══════╪══════╡
│ lo   ┆ yes  │
│ lo   ┆ yes  │
│ med  ┆ no   │
│ hi   ┆ yes  │
│ hi   ┆ no   │
└──────┴──────┘

答案4

得分: 1

你可以这样进行融合/拆分/过滤/旋转:

df \
    .with_row_count("i") \
    .melt('i') \
    .with_columns(pl.col('variable').str.split('_')) \
    .with_columns(col=pl.col('variable').arr.first(), val=pl.col('variable').arr.last()) \
    .filter(pl.col('value')==1) \
    .pivot('val','i','col') \
    .sort('i').drop('i')
英文:

You can do a melt/split/filter/pivot like this:

df \
    .with_row_count("i") \
    .melt('i') \
    .with_columns(pl.col('variable').str.split('_')) \
    .with_columns(col=pl.col('variable').arr.first(), val=pl.col('variable').arr.last()) \
    .filter(pl.col('value')==1) \
    .pivot('val','i','col') \
    .sort('i').drop('i')

huangapple
  • 本文由 发表于 2023年4月6日 18:57:48
  • 转载请务必保留本文链接:https://go.coder-hub.com/75948718.html
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