bars = exchange.fetch_ohlcv(symbol, timeframe=zamanAraligi, since=None, limit=500) df = pd.DataFrame(bars, columns=["timestamp", "open", "high", "low", "close", "volume"]) #ema control e8 = ta.ma("ema", df.close, length=8) e13 = ta.ma('ema', df.close, length=13) e21 = ta.ma('ema', df.close, length=21) e55 = ta.ma('ema', df.close, length=55) total = 0 if any(e8 > e13): total += 1 else: total -= 1 if any(e13 > e21): total += 1 else: total -= 1 if any(e21 > e55): total += 1 else: total -= 1 print("Ema 8 =", e8) print("Ema 13 =", e13) print("Ema 21 =", e21) print("Ema 55 =", e55) print("rating =", total)Of course, I'm waiting for help, sorry for the delay. I had to go out for an unexpected reason, the final code is by, if I don't use "any" it gives the error I mentioned, when I use it, it doesn't return the correct result no matter what I do.
Why is the if construct not calculating correctly?
Why is the if construct not calculating correctly?
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Mar-21-2022, 01:20 PM
(This post was last modified: Mar-21-2022, 01:21 PM by deanhystad.)
Do you know what any does?
import numpy as np a = np.array([1, 1, 1]) b = np.array([1000, 1000, 0]) print(any(a > b)) Because one element in a is greater than the corresponding element in b, any(a > b) is True. Is that the comparison you want to make?I suggested you compare if e8[-1] > e13[-1]:because you are printing out e8[-1] and e13[-1] and using that in your argument that the if statement is producing the wrong result. For some reason you instead compared: if e8 > e13:Which is an error because Python does not know how you want to compare the two arrays. I don't think comparing the very last element in the array is a good way to compare two arrays, but at least the results would align with what you think they should be. Maybe you should compare the mean? import numpy as np a = np.array([1, 1, 1]) b = np.array([1000, 1000, 0]) print("any a > b", any(a > b)) print("last a > b", a[-1] > b[-1]) print("average a > b", a.mean() > b.mean()) Or maybe there are other characteristics that are more important.
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