避免这种嵌套 for 循环的方法

Question

此代码是否需要嵌套的 for 循环，或者是否有更有效的解决方法？

这是一个简化版本，它在由 1 到 1000 的 20 个随机整数组成的数据集中搜索连续的重叠区间。它运行 1-100 的误差值，通过从 20 中添加/减去它们来创建区间随机整数。

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示例：

假设数据帧大小为 10 而不是 20 的输入：

df = [433, 3, 4, 5, 6, 7, 378, 87, 0, 500]

for 循环中 error = 1 的输出：

重叠 = {0:[[1, 2, 3, 4, 5]]}

def find_overlap(df, error):
    """
    df: dataframe with random 20 integers from 1-1000
    error: used to create the interval by +/- to each value in the dataframe
    returns: list of list of indexes overlapping
    """

    # add the interval to the dataframe as columns of minimum and maximum
    df["min"] = df["x"] - error
    df["max"] = df["x"] + error

    # overlaps stores lists of indexes that overlap
    overlaps = []

    # fill in data for start
    temporary = [0]
    minimum = df["min"].iloc[0]
    maximum = df["min"].iloc[0]

    # iterates through the dataframe checking for overlap between successive intervals
    for index , row in df.iterrows():
        current_min = row["min"]
        current_max = row["max"]

        # yes overlap
        if (current_min <= maximum) and (current_max >= minimum):
            temporary.append(index)
            if current_min > minimum:
                minimum = current_min
            if current_max < maximum:
                maximum = current_max
            continue

        # no overlap - also check for 5 successive overlaps
        if len(temporary) >= 5:
            overlaps.append(temporary)
        temporary = [index]
        minimum = current_min
        maximum = current_max

    return overlaps



# creates dataframe with 20 random integers from 1 to 1000
df = pd.DataFrame(np.random.randint(1, 1000, 20), columns=["x"])

overlaps = {}
for error in range(0,100):
    lst = find_overlap(df, error)
    if len(lst):
        overlaps[error] = lst

print(overlaps)

Answer 1

所以，根据我从你的代码中了解到的......你正在寻找：

1. 计算 x 的所有值之间的差异。
2. 确定它是否小于 error，其中 error 的范围为 [0, 100)。
3. 选择所有大小为 5 的子数组。

假设我的解释是正确的...您实际上可以将其矢量化并避免 for 循环，就像您的直觉让您相信一样。最终，如果我的解释不正确，这至少应该为您创建所需代码的矢量化版本提供一个不错的开始。 ?

更新的解决方案（考虑 5 元组）

import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.randint(1, 1000, 20), columns=["x"])

overlaps = {}

for margin in range(0, 100):
    diffs = np.abs(df["x"].values - np.roll(df["x"], margin))
    # np.convolve is analogous to a sliding window sum
    quint = np.convolve(diffs == margin, np.ones(5), "valid")
    index = np.nonzero(quint == 5)[0]
    if index.size > 0:
        overlaps[margin] = [list(range(i, i + 5)) for i in index]

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原始解决方案（不考虑 5 元组）

import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.randint(1, 1000, 20), columns=["x"])

overlaps = {}

for margin in range(0, 100):
    diffs = np.abs(df["x"].values - np.roll(df["x"], margin))
    index = np.nonzero(diff == margin)[0]
    if idx.size > 0:
        overlaps[margin] = idx

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如果您不熟悉 numpy，.size 会为您提供 ndarray 的总大小。 （因此形状为 (10, 20, 30) 的 3D 数组的大小为 6000。）