Check if a value in dataframe will increase or decrease a certain percentage

Question

I have an OHLC dataframe, eg:

index	open	close	high	low
2021-03-23 10:00:00+00:00	1421.100	1424.500	1427.720	1422.650
2021-03-23 11:00:00+00:00	1424.500	1421.480	1422.400	1411.890
2021-03-23 12:00:00+00:00	1421.480	1435.170	1443.980	1433.780
2021-03-23 13:00:00+00:00	1435.170	1440.860	1443.190	1437.590
2021-03-23 14:00:00+00:00	1440.860	1438.920	1443.570	1435.200
2021-03-23 15:00:00+00:00	1438.920	1435.990	1444.840	1435.060
2021-03-23 16:00:00+00:00	1435.990	1441.920	1446.610	1441.450

Now I want to find out, if the price will first increase or decrease eg for 1%. What I have so far is the following working code:

def check(x):

    check = ohlc[ohlc.index > x.name]
    price = ohlc.at[x.name, 'close']

    high_thr = price * 1.01
    low_thr = price * 0.99

    high_indexes = check[check['high'] > high_thr]
    low_indexes = check[check['low'] < low_thr]

    if high_indexes.shape[0] > 0 and low_indexes.shape[0] > 0:

        high = high_indexes.index[0]
        low = low_indexes.index[0]

        if high < low:
            return 1
        elif high > low:
            return -1
        else:
            return 0
    else:
        return 0


ohlc['check'] = ohlc.apply(find_threshold, axis=1)

This is extremely slow for larger datasets. Is there any other better way than iterating over every row, slicing and finding all indexes to get the nearest one?

Answer 1

I think the best way to do this is not too different from how you're doing it:

from datetime import timedelta

def check(x, change=0.01):
    time = x.name
    price = ohlc.loc[time, 'close']
    while True:
        if time not in ohlc.index:          # If we reach the end
            return 0
        high = ohlc.loc[time, 'high']
        low = ohlc.loc[time, 'low']
        if high > (1.0 + change) * price:   # Upper thresh broken
            return 1
        elif low < 1.0 - change) * price:   # Lower thresh broken
            return -1
        time = time + timedelta(hours=1)    # Time update

ohlc['check'] = ohlc.apply(check, axis=1)

If efficiency is what you're worreid about, applying this way is slightly more efficient because it only looks ahead as far as it needs to to break the threshold. Optionally, you could limit this to up to, say, 100 hours into the future by modifying the while loop, capping the number of checks per row to 100:

    endtime = time + timedelta(hours=100)
    while time < endtime:
        # etc