CODEWITHSUNDEEP
python
Evan Day
Evan Day

Reputation: 23

Python looping list

I have function 1

def predict(x, y):
    z = []
    for i in x:
        if i >= y:
            z.append(1)
        else:
            z.append(0)
    return z

Now, for each thresholds value, I want to use that and pred_probs to call my first function and then populate the empty lists. so the first call would be predict(pred_probs, .0.00). the second call in the loop would be predict(pred_probs, 0.25) and so on. each loop would append the output the lists

   pred_1 = []
    pred_2 = []
    pred_3 = []
    pred_4 = []
    pred_5 = []
    thresholds = [0.00, 0.25, 0.50, 0.75, 1.00]
    pred_probs = [0.875, 0.325, 0.6, 0.09, 0.4]

    for i in thresholds:
        pred =  predict(pred_probs,i)
        pred1.append(pred)

The desired outcome would be

pred_1 = [1,1,1,1,1]
pred_2 = [1,1,1,0,1]
pred_3 = [1,0,1,0,0]
pred_4 = [1,0,0,0,0]
pred_5 = [0,0,0,0,0]

the problem is that I'm not sure how to access the lists individually.

this is the output i receive:

[1, 1, 1, 1, 1]
[1, 1, 1, 0, 1]
[1, 0, 1, 0, 0]
[1, 0, 0, 0, 0]
[0, 0, 0, 0, 0]

however, i'm not sure how to take the first list and assign it to pred_1 and then son

Upvotes: 1

Views: 70

Answers (2)

ekrall
ekrall

Reputation: 192

here's an alternate way that vectorizes the function

import numpy as np


def predict(x, y):
    if x >= y:
        return 1
    return 0


thresholds = [0.00, 0.25, 0.50, 0.75, 1.00]
pred_probs = [0.875, 0.325, 0.6, 0.09, 0.4]

pred_lists = np.ndarray(shape=(len(thresholds), len(pred_probs)), dtype=np.int32)

v_predict = np.vectorize(predict)

for n, threshold in enumerate(thresholds):
    pred_lists[n, :] = v_predict(pred_probs, threshold)
print(pred_lists)

Output

[[1 1 1 1 1]
 [1 1 1 0 1]
 [1 0 1 0 0]
 [1 0 0 0 0]
 [0 0 0 0 0]]

Upvotes: 1

ogdenkev
ogdenkev

Reputation: 2374

Here is how you could do exactly what you asked.

for i, arr in zip(thresholds, [pred_1, pred_2, pred_3, pred_4, pred_5]):
    pred = predict(pred_probs, i)
    arr.extend(pred)

However, you may consider whether 5 lists is really what you want. It might be easier to do

pred = []
for i in thresholds:
    vals = predict(pred_probs, i)
    pred.append(vals)
print(pred)

This will give

[[1, 1, 1, 1, 1],
 [1, 1, 1, 0, 1],
 [1, 0, 1, 0, 0],
 [1, 0, 0, 0, 0],
 [0, 0, 0, 0, 0]]

Upvotes: 1

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