Question

附录一

data = {
     value : [1,9,6,7,3, 2,4,5,1,9]
}

每一行,我想找到比目前增加的最新内容的逐年增长。

因此,我的预期产出是:

[None, 0, 1, 2, 1, 1, 3, 4, 1, 0]

the first element 1 has no previous element, so I want None in the result
the next element 9 is at least as large than all its previous elements, so I want 0 in the result
the next element 6, has its previous element 9 which is larger than it. The distance between them is 1. So, I want 1 in the result here.

我知道,我可以在沙尔的 lo中这样做(如果我写延期的话,可在C/Rust)。

我的问题是:能否解决这一完全使用数据机操作? anda或极地,要么是罚款。但只有数据基操作。

因此,以下人士无一可查:

apply
map_elements
map_rows
iter_rows
Python for loops which loop over the rows and extract elements one-by-one from the dataframes

Answer 1

a. 使用pandas/a>/

# mask & indices
tril = np.tril(arr[:, None] < arr)
last = np.where(tril, ser.index, -1).max(axis=1)

# distance
dist = (ser.index - last).where(last != -1, 0).astype("Int64")
dist.array[0] = np.nan

产出:

>>> dist.tolist()

# [<NA>, 0, 1, 2, 1, 1, 3, 4, 1, 0]

Used input :

import pandas as pd
import numpy as np

ser = pd.Series(data["value"])
arr = ser.to_numpy()

Answer 2

Using janitor/code>s

import janitor

tmp = df.reset_index()

df[ out ] = (tmp
   .conditional_join(tmp, ( index ,  index ,  > ), ( value ,  value ,  < ),
                     keep= last , how= left , right_columns= index )
   [( right ,  index )]
   .rsub(tmp.index)
   .fillna(pd.Series(0, index=tmp.index[1:])).to_numpy()
)


产出:
   value  out
0      1  NaN
1      9  0.0
2      6  1.0
3      7  2.0
4      3  1.0
5      2  1.0
6      4  3.0
7      5  4.0
8      1  1.0
9      9  0.0

Answer 3

这些问题难以解决,但你可以使用 from numba import njit, prange @njit(parallel=True) def get_values(values): out = np.zeros_like(values, dtype=np.float64) for i in prange(len(values)): idx = np.int64(i) v = values[idx] while idx > -1 and values[idx] <= v: idx -= 1 if idx > -1: out[i] = i - idx out[0] = np.nan return out data = { "value": [1, 9, 6, 7, 3, 2, 4, 5, 1, 9], "out": [None, 0, 1, 2, 1, 1, 3, 4, 1, 0], } df = pd.DataFrame(data) df["out2"] = get_values(df["value"].values) print(df)

印刷:

   value  out  out2
0      1  NaN   NaN
1      9  0.0   0.0
2      6  1.0   1.0
3      7  2.0   2.0
4      3  1.0   1.0
5      2  1.0   1.0
6      4  3.0   3.0
7      5  4.0   4.0
8      1  1.0   1.0
9      9  0.0   0.0

基准(从1到100年有1 000-000件物品):

from timeit import timeit

data = {
    "value": np.random.randint(1, 100, size=1_000_000),
}
df = pd.DataFrame(data)

t = timeit( df["out"] = get_values(df["value"].values) , globals=globals(), number=1)
print(t)

我机器上的印刷(AMD 5700x):

0.3559090679627843

Answer 4

我猜测你正在寻找Rust执行算法的一部分,因此我提议如下:

import pandas as pd

data = {
     value : [1, 9, 6, 7, 3, 2, 4, 5, 1, 9]
}
df = pd.DataFrame(data)

values = df[ value ].tolist()

### Algorithm for implementation in Rust, C ...
r = []
for i in range(0, len(values)):
    curr = values[:i+1]
    prev = curr[:-1]
    last = curr[-1]
    if len(curr)>1:
        prev.reverse()
        dist=0
        for j in prev:
            if last < max(prev):
                dist+=1
            else:
                r.append(dist)
                break
            if last < j:
                r.append(dist)
                break
    else:
        r.append(None)

print(r)

[None, 0, 1, 2, 1, 1, 3, 4, 1, 0]

在Rust、Adhur或以下任何因素的计算之后,在数据框架内进行计算:

df[ out ] = r

print(df)

   value  out
0      1  NaN
1      9  0.0
2      6  1.0
3      7  2.0
4      3  1.0
5      2  1.0
6      4  3.0
7      5  4.0
8      1  1.0
9      9  0.0

Answer 5

This iterates only on the range of rows that this should look. It doesn t loop over the rows themselves in python. If your initial bound_range covers all the cases then it won t ever actually do a loop.

lb=0
bound_range=3
df=df.with_columns(z=pl.lit(None, dtype=pl.UInt64))
while True:
    df=df.with_columns(
        z=pl.when(pl.col( value )>=pl.col( value ).shift(1).cum_max())
            .then(pl.lit(0, dtype=pl.UInt64))
            .when(pl.col( z ).is_null())
            .then(
                pl.coalesce(
                    pl.when(pl.col( value )<pl.col( value ).shift(x))
                        .then(pl.lit(x, dtype=pl.UInt64))
                        for x in range(lb, lb+bound_range)
                )
            )
            .otherwise(pl.col( z ))
            )
    if df[1:][ z ].drop_nulls().shape[0]==df.shape[0]-1:
        break
    lb+=bound_range

例如,我制定了<条码>有条不紊的<>条/条码>至3条,以确保至少休息一次。我带着1M随机随机分类的0至9(含括)进行处理,我定有约束——安排到50,并在2个秘密下进行。你们可以通过检查检查,在休息室之间划出这一条码。

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