因为你在回顾之前的值,看看是否有连续的值,你将不得不以某种方式与索引交互。该解决方案首先查看当前索引中该值之前的任何值,以查看它们是否小于或大于该值,然后将任何值设置为False,其中后面有False。它还避免了在数据帧上创建迭代器,这可能会加快大型数据集的操作。

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)

   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

import pandas as pd
import numpy as np

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



Lower=[(value[x]<value[:x]).sum() for x in range(len(value))]
Higher=[(value[x]>value[:x]).sum() for x in range(len(value))]


df=pd.DataFrame({"value":value,"Higher":Higher,"Lower":Lower})

print(df)





      Lower  Higher  value
0       0      0     30
1       1      0     10
2       1      1     20
3       1      2     25
4       0      4     35
5       4      1     15

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

a_not_done_greater = pd.Series(np.ones(len(a))).astype(bool)
a_not_done_less = pd.Series(np.ones(len(a))).astype(bool)

a_streak_greater = pd.Series(np.zeros(len(a))).astype(int)
a_streak_less = pd.Series(np.zeros(len(a))).astype(int)

s = 1
not_done_greater = True
not_done_less = True

while not_done_greater or not_done_less:
    if not_done_greater:
        a_greater_than_shift = (a > a.shift(s))
        a_streak_greater = a_streak_greater + (a_not_done_greater.astype(int) * a_greater_than_shift)
        a_not_done_greater = a_not_done_greater & a_greater_than_shift
        not_done_greater = a_not_done_greater.any()

    if not_done_less:
        a_less_than_shift = (a < a.shift(s))
        a_streak_less = a_streak_less + (a_not_done_less.astype(int) * a_less_than_shift)
        a_not_done_less = a_not_done_less & a_less_than_shift
        not_done_less = a_not_done_less.any()

    s = s + 1


res = pd.concat([a, a_streak_greater, a_streak_less], axis=1)
res.columns = ['value', 'greater_than_streak', 'less_than_streak']
print(res)

给出了数据帧

   value  greater_than_streak  less_than_streak
0     30                    0                 0
1     10                    0                 1
2     20                    1                 0
3     25                    2                 0
4     35                    4                 0
5     15                    0                 3
6     15                    0                 0

计算连续值。我无法想出一个可行的解决方案,所以我们回到循环。

df = pd.Series(np.random.rand(10000))

def count_bigger_consecutives(values):
  length = len(values)
  result = np.zeros(length)
  for i in range(length):
    for j in range(i):
      if(values[i]>values[j]):
        result[i] += 1
      else:
        break
  return result

%timeit count_bigger_consecutives(df.values)
1 loop, best of 3: 365 ms per loop

如果您担心性能问题,可以使用 numba

from numba import jit 
@jit(nopython=True)
def numba_count_bigger_consecutives(values):
  length = len(values)
  result = np.zeros(length)
  for i in range(length):
    for j in range(i):
      if(values[i]>values[j]):
        result[i] += 1
      else:
        break
  return result

%timeit numba_count_bigger_consecutives(df.values)
The slowest run took 543.09 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 161 µs per loop

下面是一位同事提出的解决方案(可能不是最有效的,但它做到了):

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

创建“更高”列

b = []

for idx, value in enumerate(a):
    count = 0
    for i in range(idx, 0, -1):
        if value < a.loc[i-1]:
            break
        count += 1
    b.append([value, count])

higher = pd.DataFrame(b, columns=['Value', 'Higher'])

c = []

for idx, value in enumerate(a):
    count = 0
    for i in range(idx, 0, -1):
        if value > a.loc[i-1]:
            break
        count += 1
    c.append([value, count])

lower = pd.DataFrame(c, columns=['Value', 'Lower'])

合并两个新系列

print(pd.merge(higher, lower, on='Value'))

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