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极地:是否有程序确保零件的发送时间为零?
原标题:Polars: Is there a procedure that ensures zero copy when calling df.to_numpy()?

For some time now I am failing to call df.to_numpy(allow_copy = True). Is there a procedure that transforms any given dataset into "zero_copy suitable" one?

对类似清单的数值 审判

data = pl.DataFrame(dict(
    points=np.random.sample((N, 3)),
    color=np.random.sample((N, 4))
),
schema=dict(
    points=pl.Array(pl.Float64, 3),
    color=pl.Array(pl.Float64, 4),
))

或简单地说,expr.cast(pl.Array(pl.Float32, 4), as recommended here。 它用于我的一个数据集,但在不同的数据集中却失败。

<rechunk(>,无任何价值和(或)具体说明order = “c”fortran>似乎也没有任何效果。

这是我先前的。 这或许过于具体,难以找到真正的答案。

问题回答

无,这项行动将照此办理。 单项分配中分配的均匀分布。 极地分给每一栏。 这意味着,如果记忆由极地分配,则不能将这种记忆转至极值2D矩阵。

极地一栏和极地轨道类型可移至零件。

Moving data from numpy to Polars and back.

2D基体按平价计价,可向极地转移零件,再将零件退回到 n。 这样做是因为原来的消费分配不会改变。 所有极地一栏都显示为平流毗连阵列的记忆。

Why only F-order?

由于极地是一栏电梯发动机,我们只储存数据。 否则,我们就需要跳上一栏的透镜。 这将重复所有法典,或增加间接的模拟,这种推论将大大放慢。

除此以外,您现在在切合线上拥有<代码>row的数值,使所有列算法都大大放慢。

C-contiguous

在毗连区,增长值回落。 如果我们需要<代码>column_ Value 我们必须使用<代码>n

[column_value_1, row_x, row_x, ...,  row_x, column_value_2]

F-contiguous

In f-contiguous, column slots are back to back, leading to fast cache efficient reading

[column_value_1, column_value_2, ..., column_value_n]

因此,“0-copy”C-contiguous dataDataFrames涉及大多数业务(即列车)的履约费用,如果lying列算法预期有数据回馈(即斜线)。

TLDR;

用于支持单体数据零件的数据,如果算法预期有切片,则以切碎或含蓄拷贝的价格支付。





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