A call to functools.reduce returns only the final result:
>>> from functools import reduce
>>> a = [1, 2, 3, 4, 5]
>>> f = lambda x, y: x + y
>>> reduce(f, a)
15
Instead of writing a loop myself, does a function exist which returns the intermediary values, too?
[3, 6, 10, 15]
(This is only a simple example, I'm not trying to calculate the cumulative sum - the solution should work for arbitrary a and f.)
You can use itertools.accumulate():
>>> from itertools import accumulate
>>> list(accumulate([1, 2, 3, 4, 5], lambda x, y: x+y))[1:]
[3, 6, 10, 15]
Note that the order of parameters is switched relative to functools.reduce().
Also, the default func (the second parameter) is a sum (like operator.add), so in your case, it's technically optional:
>>> list(accumulate([1, 2, 3, 4, 5]))[1:] # default func: sum
[3, 6, 10, 15]
And finally, it's worth noting that accumulate() will include the first term in the sequence, hence why the result is indexed from [1:] above.
In your edit, you noted that...
This is only a simple example, I'm not trying to calculate the cumulative sum - the solution should work for arbitrary a and f.
The nice thing about accumulate() is that it is flexible about the callable it will take. It only demands a callable that is a function of two parameters.
For instance, builtin max() satisfies that:
>>> list(accumulate([1, 10, 4, 2, 17], max))
[1, 10, 10, 10, 17]
This is a longer form of using the unnecessary lambda:
>>> # Don't do this
>>> list(accumulate([1, 10, 4, 2, 17], lambda x, y: max(x, y)))
[1, 10, 10, 10, 17]
import numpy as np
x=[1, 2, 3, 4, 5]
y=np.cumsum(x) # gets you the cumulative sum
y=list(y[1:]) # remove the first number
print(y)
#[3, 6, 10, 15]
