Clash Royale CLAN TAG#URR8PPP

Computing sum of consecutive values in a vector that are greater than a constant number?

I couldn't summarize my question in the title very well. I'm writing a code and in one part of the code I need to compute the following:

Let's say we have a vector (e.g. a numpy array):

a = [3.2, 4, 7, 2, 8, 9, 7, 1.7, 2, 8, 9, 1, 3]


We want to turn any number greater than 5 to 5:

a = [3.2, 4, 5, 2, 5, 5, 5, 1.7, 2, 5, 5, 1, 3]


Then we compute the sum of consecutive 5s and the number that follows them and replace all these elements with the resulting sum:

a = [3.2, 4, 5+ 2, 5+ 5+ 5+ 1.7, 2, 5+ 5+ 1, 3]


so the resulting array would be:

a = [3.2, 4, 7, 16.7, 2, 11, 3]


I can do this using a for loop like this:

indx = np.where(a>5)[0]
a[indx] = 5
counter = 0
c =
while (counter < len(a)):
elem = a[counter]
if elem ~= 5:
c.append(elem)
else:
temp = 0
while(elem==5):
temp += elem
counter +=1
elem = a[counter]
temp += elem
c.append(temp)
counter += 1


Is there a way to avoid using the for loop? Perhaps by using the indx variable?

I have a vague idea if we turn it into a string:
a = '[3.2, 4, 5, 2, 5, 5, 5, 1.7, 2, 5, 5, 1, 3]'
and then change anywhere we have ' 5,' with ' 5+' and then use eval(a). However, is there an efficient way to find all indices containing a sub-string? How about the fact that strings are immutable?

' 5,'

' 5+'

eval(a)

np.where(a>=5)

3 Answers
3

You can use pandas for data manipulation, using cumsum and shift to groupby your values with your logic, and aggregating it with sum

pandas

cumsum

shift

groupby

sum

df = pd.DataFrame(a, columns=['col1'])
df.loc[df.col1 > 5] = 5
s = df.col1.groupby((df.col1 != 5).cumsum().shift().fillna(0)).sum()

col1
0.0 3.2
1.0 4.0
2.0 7.0
3.0 16.7
4.0 2.0
5.0 11.0
6.0 3.0


To get a numpy back, just get .values

.values

>>> s.values
array([ 3.2, 4. , 7. , 16.7, 2. , 11. , 3. ])


This is what you want (all in vectorized numpy):

import numpy as np

a = np.array([0, 3.2, 4, 7, 2, 8, 9, 7, 1.7, 2, 8, 9, 1, 3, 0]) # add a 0 at the beginning and the end
aa = np.where(a>5, 5, a) # clip values to 5, can use np.clip(a, None, 5) too...
c = np.cumsum(aa) # get cumulative sum
np.diff(c[aa < 5]) # only keep values where original array is less than 5, then diff again

array([ 3.2, 4. , 7. , 16.7, 2. , 11. , 3. ])


arr = [3.2, 4, 7, 2, 8, 9, 7, 1.7, 2, 8, 9, 1, 3,5,2,5,5,5,5,5,5,5,5,5,5,10]

5

numpy.clip

I think you can do this in a single pass. For each item:

.

a = [3.2, 4, 7, 2, 8, 9, 7, 1.7, 2, 8, 9, 1, 3]
result =

current_sum = 0
for item in a:
if item < 5:
result.append(current_sum + item)
current_sum = 0
else:
current_sum += 5

if current_sum:
result.append(current_sum)

>>> result
[3.2, 4, 7, 16.7, 2, 11, 3]


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.