python - pandas - 计算高于/低于当前行的连续值

Question

我正在寻找一种获取 Pandas 系列并返回新系列的方法，该新系列表示高于/低于系列中每一行的先前连续值的数量:

a = pd.Series([30, 10, 20, 25, 35, 15])

...应该输出:

Value   Higher than streak  Lower than streak
30      0                   0
10      0                   1
20      1                   0
25      2                   0
35      4                   0
15      0                   3

这将使某人能够确定每个“区域最大/最小”值在时间序列中的重要性。

提前致谢。

Answer 1

由于您要向后查看以前的值以查看是否存在连续值，因此您将不得不以某种方式与索引进行交互。此解决方案首先查看当前索引处的值之前的任何值，以查看它们是小于还是大于该值，然后将任何值设置为 False(如果后面有 False)。它还避免在 DataFrame 上创建迭代器，这可能会加速更大数据集的操作。

import pandas as pd
from operator import gt, lt

a = pd.Series([30, 10, 20, 25, 35, 15])

def consecutive_run(op, ser, i):
    """
    Sum the uninterrupted consecutive runs at index i in the series where the previous data
    was true according to the operator.
    """
    thresh_all = op(ser[:i], ser[i])
    # find any data where the operator was not passing.  set the previous data to all falses
    non_passing = thresh_all[~thresh_all]
    start_idx = 0
    if not non_passing.empty:
        # if there was a failure, there was a break in the consecutive truth values,
        # so get the final False position. Starting index will be False, but it
        # will either be at the end of the series selection and will sum to zero
        # or will be followed by all successive True values afterwards
        start_idx = non_passing.index[-1]
    # count the consecutive runs by summing from the start index onwards
    return thresh_all[start_idx:].sum()


res = pd.concat([a, a.index.to_series().map(lambda i: consecutive_run(gt, a, i)),
                 a.index.to_series().map(lambda i: consecutive_run(lt, a, i))],
       axis=1)
res.columns = ['Value', 'Higher than streak', 'Lower than streak']
print(res)

结果:

   Value  Higher than streak  Lower than streak
0     30                   0                  0
1     10                   1                  0
2     20                   0                  1
3     25                   0                  2
4     35                   0                  4
5     15                   3                  0