Python 在熊猫中，我如何计算连续的积极和消极因素？

在python或numpy中，是否有一个内置函数或函数组合可以计算一行中正值或负值的数量

这可以被认为类似于一个轮盘赌轮盘，轮盘上有一排黑色或红色的数字

输入系列数据示例：

Date
2000-01-07    -3.550049
2000-01-10    28.609863
2000-01-11    -2.189941
2000-01-12     4.419922
2000-01-13    17.690185
2000-01-14    41.219971
2000-01-18     0.000000
2000-01-19   -16.330078
2000-01-20     7.950195
2000-01-21     0.000000
2000-01-24    38.370117
2000-01-25     6.060059
2000-01-26     3.579834
2000-01-27     7.669922
2000-01-28     2.739991
2000-01-31    -8.039795
2000-02-01    10.239990
2000-02-02    -1.580078
2000-02-03     1.669922
2000-02-04     7.440186
2000-02-07    -0.940185

期望输出：

-  in a row 5 times
+  in a row 4 times
++  in a row once
++++  in a row once
+++++++ in a row once

您可以使用函数

import itertools

l = [-3.550049, 28.609863, -2.189941,  4.419922, 17.690185, 41.219971,  0.000000, -16.330078,  7.950195,  0.000000, 38.370117,  6.060059,  3.579834,  7.669922,  2.739991, -8.039795, 10.239990, -1.580078,  1.669922,  7.440186, -0.940185]

r_pos = {}
r_neg = {}
for k, v in itertools.groupby(l, lambda e:e>0):
    count = len(list(v))
    r = r_pos
    if k == False:
        r = r_neg
    if count not in r.keys():
        r[count] = 0
    r[count] += 1

for k, v in r_neg.items():
    print '%s in a row %s time(s)' % ('-'*k, v)

for k, v in r_pos.items():
    print '%s in a row %s time(s)' % ('+'*k, v)

输出

- in a row 6 time(s)
+ in a row 2 time(s)
++ in a row 1 time(s)
++++ in a row 1 time(s)
+++++++ in a row 1 time(s)

根据你认为的正值，你可以改变线<代码> lambda E:E＞0 /代码>

< P>非负：

from functools import reduce  # For Python 3.x
ser = df['x'] >= 0
c = ser.expanding().apply(lambda r: reduce(lambda x, y: x + 1 if y else x * y, r))
c[ser & (ser != ser.shift(-1))].value_counts()
Out: 
1.0    2
7.0    1
4.0    1
2.0    1
Name: x, dtype: int64

负片：

ser = df['x'] < 0
c = ser.expanding().apply(lambda r: reduce(lambda x, y: x + 1 if y else x * y, r))
c[ser & (ser != ser.shift(-1))].value_counts()

Out: 
1.0    6
Name: x, dtype: int64

---

现在，为了确定转折点，条件是当前值与下一个值不同，并且是真的。如果你选择了这些，你就有了计数。

到目前为止，这就是我想到的，它工作并输出一行中每个负值、正值和零值出现多少次的计数。也许有人可以使用上面ayhan和Ghilas发布的一些建议使它更简洁

from collections import Counter

ser = [-3.550049, 28.609863, -2.1, 89941,4.419922,17.690185,41.219971,0.000000,-16.330078,7.950195,0.000000,38.370117,6.060059,3.579834,7.669922,2.739991,-8.039795,10.239990,-1.580078, 1.669922, 7.440186,-0.940185]

c = 0
zeros, neg_counts, pos_counts = [], [], []
for i in range(len(ser)):
    c+=1
    s = np.sign(ser[i])
    try:
        if s != np.sign(ser[i+1]):
            if s == 0:
               zeros.append(c)
            elif s == -1:
                neg_counts.append(c)
            elif s == 1:
                pos_counts.append(c)
            c = 0
    except IndexError:
        pos_counts.append(c) if s == 1 else neg_counts.append(c) if s ==-1 else zeros.append(c)

print(Counter(neg_counts))
print(Counter(pos_counts))
print(Counter(zeros))

输出：

---

当然，有

cumcount（）

function您所说的

一行中连续的正数和负数是什么意思？你能给我们一个样品箱吗？谢谢你的帮助。我没能让你的建议起作用，但把你的答案和下面Ghilas的答案结合起来，我就能找到一些适合我需要的东西。再次感谢！谢谢你的建议。把这个和前面的答案结合起来，我就能够把我需要的东西放在一起。令人惊叹的！
from collections import Counter

ser = [-3.550049, 28.609863, -2.1, 89941,4.419922,17.690185,41.219971,0.000000,-16.330078,7.950195,0.000000,38.370117,6.060059,3.579834,7.669922,2.739991,-8.039795,10.239990,-1.580078, 1.669922, 7.440186,-0.940185]

c = 0
zeros, neg_counts, pos_counts = [], [], []
for i in range(len(ser)):
    c+=1
    s = np.sign(ser[i])
    try:
        if s != np.sign(ser[i+1]):
            if s == 0:
               zeros.append(c)
            elif s == -1:
                neg_counts.append(c)
            elif s == 1:
                pos_counts.append(c)
            c = 0
    except IndexError:
        pos_counts.append(c) if s == 1 else neg_counts.append(c) if s ==-1 else zeros.append(c)

print(Counter(neg_counts))
print(Counter(pos_counts))
print(Counter(zeros))

Counter({1: 5})
Counter({1: 3, 2: 1, 4: 1, 5: 1})
Counter({1: 2})