CODEWITHSUNDEEP
pythonpandas
xyhuang
xyhuang

Reputation: 424

How to calculate continuous increase count number for pandas Series?

I want to calculate the future continuous increase or decrease count for each element.

Assume I have a series:

l=[10, 9, 8, 9, 10, 9, 10, 11, 10, 12, 10]
s=pd.Series(l)

The expected output is:

o=[-2, -1, 2, 1, -1, 2, 1, -1, 1, -1, 0]

Here is the explanation:

the first element is 10, the next is 9, the next is 8, so, it decrease 2 times, so the output is -2, for the third element is 8, it increase 2 times (the next is 9, 10), so output is 2.

The key point is continuous, once it has reverse data, it should be stop look forward.

How can I implement that? I think for loop is not good enough

Upvotes: 0

Views: 451

Answers (1)

Pygirl
Pygirl

Reputation: 13339

1st get the sign of difference. Then find how many are consecutive same sign. If same then add the quantity. try:

import numpy as np
l=np.array([10, 9, 8, 9, 10, 9, 10, 11, 10, 12, 10])
sign = np.sign(np.diff(l))
count = {}
for i in range(len(sign)-1):
    count[i] = sign[i]
    for j in range(i+1, len(sign)):
        if sign[i]==sign[j]:
            count[i] += sign[j]
        else:
            break
count[i+1] = 0
res = count.values()

res:

dict_values([-2, -1, 2, 1, -1, 2, 1, -1, 1, 0])

Edit:

If you want to use pandas

df = pd.DataFrame(sign, columns=['col'])
df = df[::-1]
x = df.groupby((df['col'] != df['col'].shift(1)).cumsum()).cumsum()['col']
x.iloc[0]=0
x = x.values[::-1]

x:

array([-2, -1,  2,  1, -1,  2,  1, -1,  1,  0])

instead of :

(df['col'] != df['col'].shift(1)).cumsum()

you can use

df.col.diff().ne(0).cumsum()

Explanation

Upvotes: 1

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