英文:
Reference polars.DataFrame.height in with_columns
问题
在这个例子中，numpy.random.randint(10, 99, 6) 中的 6 是硬编码的DataFrame的高度，所以如果我将间隔从 8h 更改为 4h（需要将 6 更改为 12），它将无法工作。
我知道可以通过中断链来实现：
df = polars.DataFrame(dict(j=polars.date_range(datetime.date(2023, 1, 1), datetime.date(2023, 1, 3), '4h', closed='left', eager=True),))df = df.with_columns(k=polars.lit(numpy.random.randint(10, 99, df.height)),)
是否有办法在一个链式表达式中实现（即引用 df.height 或等效的内容）？
英文:
Take this example:
df = (polars.DataFrame(dict(j=polars.date_range(datetime.date(2023, 1, 1), datetime.date(2023, 1, 3), '8h', closed='left', eager=True),)).with_columns(k=polars.lit(numpy.random.randint(10, 99, 6)),))j k2023-01-01 00:00:00 472023-01-01 08:00:00 222023-01-01 16:00:00 822023-01-02 00:00:00 192023-01-02 08:00:00 852023-01-02 16:00:00 15shape: (6, 2)
Here, numpy.random.randint(10, 99, 6) uses hard-coded 6 as the height of DataFrame, so it won't work if I changed e.g. the interval from 8h to 4h (which would require changing 6 to 12).
I know I can do it by breaking the chain:
df = polars.DataFrame(dict(j=polars.date_range(datetime.date(2023, 1, 1), datetime.date(2023, 1, 3), '4h', closed='left', eager=True),))df = df.with_columns(k=polars.lit(numpy.random.randint(10, 99, df.height)),)j k2023-01-01 00:00:00 472023-01-01 04:00:00 222023-01-01 08:00:00 822023-01-01 12:00:00 192023-01-01 16:00:00 852023-01-01 20:00:00 152023-01-02 00:00:00 892023-01-02 04:00:00 742023-01-02 08:00:00 262023-01-02 12:00:00 112023-01-02 16:00:00 862023-01-02 20:00:00 81shape: (12, 2)
Is there a way to do it (i.e. reference df.height or an equivalent) in one chained expression though?
答案1
得分: 2
你可以使用 .pipe()
(pl.date_range(datetime.date(2023, 1, 1),datetime.date(2023, 1, 3),'4h',closed='left',eager=True).to_frame().pipe(lambda df:df.with_columns(rand =pl.lit(np.random.randint(10, 99, df.height)))))
形状: (12, 2)┌─────────────────────┬──────┐│ date ┆ rand ││ --- ┆ --- ││ datetime[μs] ┆ i64 │╞═════════════════════╪══════╡│ 2023-01-01 00:00:00 ┆ 39 ││ 2023-01-01 04:00:00 ┆ 45 ││ 2023-01-01 08:00:00 ┆ 95 ││ 2023-01-01 12:00:00 ┆ 72 ││ … ┆ … ││ 2023-01-02 08:00:00 ┆ 34 ││ 2023-01-02 12:00:00 ┆ 42 ││ 2023-01-02 16:00:00 ┆ 30 ││ 2023-01-02 20:00:00 ┆ 83 │└─────────────────────┴──────┘
英文:
You can use .pipe()
(pl.date_range(datetime.date(2023, 1, 1),datetime.date(2023, 1, 3),'4h',closed='left',eager=True).to_frame().pipe(lambda df:df.with_columns(rand =pl.lit(np.random.randint(10, 99, df.height)))))
shape: (12, 2)┌─────────────────────┬──────┐│ date ┆ rand ││ --- ┆ --- ││ datetime[μs] ┆ i64 │╞═════════════════════╪══════╡│ 2023-01-01 00:00:00 ┆ 39 ││ 2023-01-01 04:00:00 ┆ 45 ││ 2023-01-01 08:00:00 ┆ 95 ││ 2023-01-01 12:00:00 ┆ 72 ││ … ┆ … ││ 2023-01-02 08:00:00 ┆ 34 ││ 2023-01-02 12:00:00 ┆ 42 ││ 2023-01-02 16:00:00 ┆ 30 ││ 2023-01-02 20:00:00 ┆ 83 │└─────────────────────┴──────┘
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