使用 Numpy argmax 来计数 vs for 循环

Question

I currently use something like the similar bit of code to determine comparison

list_of_numbers = [29800.0, 29795.0, 29795.0, 29740.0, 29755.0, 29745.0]
high = 29980.0
lookback = 10
counter = 1

for number in list_of_numbers:
    if (high >= number) \
    and (counter < lookback):
        counter += 1
    else:
        break

this will give you the output 7 and functions as you expect it to. however is very taxing on large data arrays. So I looked for a solution and came up with np.argmax() but there seems to be an issue. For example the following:

list_of_numbers = [29800.0, 29795.0, 29795.0, 29740.0, 29755.0, 29745.0]
np_list = np.array(list_of_numbers)
high = 29980.0

print(np.argmax(np_list > high) + 1)

this will output 1, just like armax is suppose to.. but I want it to output 7. Is there another method to do this that will give me similar output to the if statement?

Answer 1

You can get a boolean array for where high >= number using NumPy, and then finding where is the first False argument in that to satisfy break condition in your code.:

boolean_arr = np.less_equal(np.array(list_of_numbers), high)
break_arr = np.where(boolean_arr == False)[0][0] + 1

To consider countering, you can use np.cumsum on the boolean array and find the first argument that satisfying specified lookback magnitude. So, the result will be smaller value between break_arr and lookback_lim :

lookback_lim = np.where(np.cumsum(boolean_arr) == lookback)[0][0]
result = min(break_arr, lookback_lim)

Answer 2

You have you less than sign backwards, no? The following should word the same as your for-loop:

print(np.min([np.sum(np_list < high)+1, lookback]))

Answer 3

A look back can be accomplished using shift. A cumcount can be used to get a running total. A query can be used as a filter