Question

我有一个指数和值介于0和1之间的Pandas Dataframe，如下所示：

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

我想检索超过5个连续值的区域的起点和终点的元组，这些值超过某个阈值（例如0.5）。所以我会有这样的事情：

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

如果第一个元组是从索引150开始的区域，则有35个值在行中都高于0.5，并且在索引185不包含时结束。

我开始只过滤0.5以上的值，如此

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

现在我有这样的价值观：

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

我无法显示我的实际数据集，但以下内容应该是一个很好的表示

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

得到以下特性：

 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
...

区域（2,4）有两个高于0.5的值。 然而，这将太短。另一方面，连续超过0.5的19个值的区域（25,44）将被添加到列表中。

Answer 1

通过查看序列和1行移位值，可以找到每个连续区域的第一个和最后一个元素，然后过滤彼此充分分离的对：

# 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]

所以例如第一个区域是：

>>> 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

Answer 2

我认为这会打印你想要的东西。它主要基于Joe Kington's answer here我认为适合对其进行投票。

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()

如果没有更优雅的方式，我会感到惊讶

在Pandas DataFrame中定义具有高于特定阈值的值的连续区域

2 个答案: