在Pandas DataFrame中界定超过某个阈值的连续区域

10 投票
2 回答
4586 浏览
提问于 2025-04-18 10:11

我有一个Pandas数据框,里面有一些索引和介于0到1之间的数值,像这样:

 6  0.047033
 7  0.047650
 8  0.054067
 9  0.064767
10  0.073183
11  0.077950

我想要找出那些连续超过5个值都高于某个阈值(比如0.5)的区域的起始和结束点的元组。这样我就能得到类似下面的结果:

 [(150, 185), (632, 680), (1500,1870)]

第一个元组表示一个区域,它从索引150开始,有35个值连续都高于0.5,并且在索引185结束(不包括185)。

我开始时只筛选出高于0.5的值,像这样:

 df = df[df['values'] >= 0.5]

现在我得到了这样的值:

632  0.545700
633  0.574983
634  0.572083
635  0.595500
636  0.632033
637  0.657617
638  0.643300
639  0.646283

我不能展示我的实际数据集,但下面这个数据集应该能很好地代表我的情况:

import numpy as np
from pandas import *

np.random.seed(seed=901212)

df = DataFrame(range(1,501), columns=['indices'])
df['values'] = np.random.rand(500)*.5 + .35

结果是:

 1  0.491233
 2  0.538596
 3  0.516740
 4  0.381134
 5  0.670157
 6  0.846366
 7  0.495554
 8  0.436044
 9  0.695597
10  0.826591
...

其中区域(2,4)有两个值高于0.5。不过这个区域太短了。 另一方面,区域(25,44)有19个值连续高于0.5,所以这个区域会被加入到列表中。

2 个回答

1

我觉得这个代码能打印出你想要的结果。它主要是参考了Joe Kington在这里的回答,我觉得给他点赞是合适的。

import numpy as np

# from Joe Kington's answer here https://stackoverflow.com/a/4495197/3751373
# with minor edits
def contiguous_regions(condition):
    """Finds contiguous True regions of the boolean array "condition". Returns
    a 2D array where the first column is the start index of the region and the
    second column is the end index."""

    # Find the indicies of changes in "condition"
    d = np.diff(condition,n=1, axis=0)
    idx, _ = d.nonzero() 

    # We need to start things after the change in "condition". Therefore, 
    # we'll shift the index by 1 to the right. -JK
    # LB this copy to increment is horrible but I get 
    # ValueError: output array is read-only without it 

    mutable_idx = np.array(idx)
    mutable_idx +=  1
    idx = mutable_idx

    if condition[0]:
        # If the start of condition is True prepend a 0
        idx = np.r_[0, idx]

    if condition[-1]:
        # If the end of condition is True, append the length of the array
        idx = np.r_[idx, condition.size] # Edit

    # Reshape the result into two columns
    idx.shape = (-1,2)
    return idx

def main():
    import pandas as pd
    RUN_LENGTH_THRESHOLD = 5
    VALUE_THRESHOLD = 0.5

    np.random.seed(seed=901212)
    data = np.random.rand(500)*.5 + .35

    df = pd.DataFrame(data=data,columns=['values'])

    match_bools =  df.values > VALUE_THRESHOLD 


    print('with boolian array')
    for start, stop in contiguous_regions(match_bools):
        if (stop - start > RUN_LENGTH_THRESHOLD):
            print (start, stop)



if __name__ == '__main__':
    main()

如果有更优雅的方法我会感到很惊讶。

24

你可以通过查看一系列数据和向右移动一行的数据,来找到每个连续区域的第一个和最后一个元素。然后再筛选出那些相隔足够远的元素对。

# tag rows based on the threshold
df['tag'] = df['values'] > .5

# first row is a True preceded by a False
fst = df.index[df['tag'] & ~ df['tag'].shift(1).fillna(False)]

# last row is a True followed by a False
lst = df.index[df['tag'] & ~ df['tag'].shift(-1).fillna(False)]

# filter those which are adequately apart
pr = [(i, j) for i, j in zip(fst, lst) if j > i + 4]

举个例子,第一个区域会是:

>>> i, j = pr[0]
>>> df.loc[i:j]
    indices    values   tag
15       16  0.639992  True
16       17  0.593427  True
17       18  0.810888  True
18       19  0.596243  True
19       20  0.812684  True
20       21  0.617945  True

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