如何将两个DataFrames合并为不同的列/大小

Question

寻找将两个DataFrames组合在一起的方法。

df1:

shape: (2, 2)
┌────────┬──────────────────────┐
│ Fruit  ┆ Phosphorus (mg/100g) │
│ ---    ┆ ---                  │
│ str    ┆ i32                  │
╞════════╪══════════════════════╡
│ Apple  ┆ 11                   │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Banana ┆ 22                   │
└────────┴──────────────────────┘

df2:

shape: (1, 3)
┌──────┬─────────────────────┬──────────────────────┐
│ Name ┆ Potassium (mg/100g) ┆ Phosphorus (mg/100g) │
│ ---  ┆ ---                 ┆ ---                  │
│ str  ┆ i32                 ┆ i32                  │
╞══════╪═════════════════════╪══════════════════════╡
│ Pear ┆ 115                 ┆ 12                   │
└──────┴─────────────────────┴──────────────────────┘

的结果应该是：

shape: (3, 3)
+--------+----------------------+---------------------+
| Fruit  | Phosphorus (mg/100g) | Potassium (mg/100g) |
| ---    | ---                  | ---                 |
| str    | i32                  | i32                 |
+========+======================+=====================+
| Apple  | 11                   | null                |
+--------+----------------------+---------------------+
| Banana | 22                   | null                |
+--------+----------------------+---------------------+
| Pear   | 12                   | 115                 |
+--------+----------------------+---------------------+

下面是我要做的代码片段：

use polars::prelude::*;

fn main() {
    let df1: DataFrame = df!("Fruit" => &["Apple", "Banana"],
                         "Phosphorus (mg/100g)" => &[11, 22])
    .unwrap();

    let df2: DataFrame = df!("Name" => &["Pear"],
                            "Potassium (mg/100g)" => &[115],
                            "Phosphorus (mg/100g)" => &[12]
    )
    .unwrap();

    let df3: DataFrame = df1
        .join(&df2, ["Fruit"], ["Name"], JoinType::Left, None)
        .unwrap();

    assert_eq!(df3.shape(), (3, 3));
    println!("{}", df3);
}

我要找的是全外连接 .

我得到的错误：

线程'main‘惊慌于’断言失败：(left == right)左：(2, 4)，右：(3, 3)，src\main.rs:18:5

Answer 1

感谢@Ayaz :)我能够制作一个通用版本，在这个版本中，我不需要每次指定共享列名。

下面是两个DataFrames的完全外部连接的版本：

use polars::prelude::*;
use array_tool::vec::{Intersect};

fn concat_df(df1: &DataFrame, df2: &DataFrame) -> Result<DataFrame, PolarsError> {
    if df1.is_empty() {
        return Ok(df2.clone());
    }

    let df1_column_names = df1.get_column_names();
    let df2_column_names = df2.get_column_names();

    let common_column_names = &df1_column_names.intersect(df2_column_names)[..];

    df1.join(
        df2,
        common_column_names,
        common_column_names,
        JoinType::Outer,
        None,
    )
}

Answer 2

您需要显式指定要合并的列，并将JoinType::Outer用于外部联接：

use polars::prelude::*;

fn main() {
    let df1: DataFrame = df!("Fruit" => &["Apple", "Banana"],
                         "Phosphorus (mg/100g)" => &[11, 22])
    .unwrap();

    let df2: DataFrame = df!("Name" => &["Pear"],
                            "Potassium (mg/100g)" => &[115],
                            "Phosphorus (mg/100g)" => &[12]
    )
    .unwrap();

    let df3: DataFrame = df1
        .join(
            &df2,
            ["Fruit", "Phosphorus (mg/100g)"],
            ["Name", "Phosphorus (mg/100g)"],
            JoinType::Outer,
            None).unwrap();

    assert_eq!(df3.shape(), (3, 3));
    println!("{}", df3);
}