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The function below uses slices to get the maximum value between 2 indexes at once. So it gets the maximum value between 0 and 10 and then 10 and 12 and such. The function is derived from the answer to this post post. Is there a way I could could replace the list comprehensions in the form of a pandas function like pd.Series(). Or if possible do it as a numpy function.

list_ = np.array([9887.89, 9902.99, 9902.99, 9910.23, 9920.79, 9911.34, 9920.01, 9927.51, 9932.3, 9932.33, 9928.87, 9929.22, 9929.22, 9935.24, 9935.24, 9935.26, 9935.26, 9935.68, 9935.68, 9940.5])
indexes = np.array([0,10,12,14])
chunks = np.split(list_, indexes[1:-1])
MAX=([c.max() for c in chunks])
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  • 1
    Why do you want to do this? Is it for performance reasons or do you need results in the form of a numpy array? Commented Jan 29, 2021 at 0:09
  • df.iloc[0:10].max() just replace 0/10 with what you want Commented Jan 29, 2021 at 0:11

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I would recommend this:

MAX = np.maximum.reduceat(list_,indexes[:-1])

output:

array([9932.33, 9929.22, 9940.5 ])

Another way that will NOT increase your performance and is merely a replacement for list comprehension in your answer (in fact, might even be slightly slower):

max = np.vectorize(np.max)
MAX = max(chunks)
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4 Comments

Is there a way I could get the indexes of 9932.33, 9929.22, 9940.5 as well. I tried to implement argmax to it howwver it didnt workout
@tonyselcuk Quite a different question but does this post answer your question stackoverflow.com/a/41835843/4975981?
@tonyselcuk Glad it resolved. Please feel free to accept the answer to close the question. Thank you.
Sorry about the delay just did it now.

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