x (> if bool else <) y

Question

I'm looking for a logic to use a function with different operators.

Is there a way to implement the logic to use a boolean to determine the operator in the comparison? something along the lines of

while df["column"][i + period] (> if bool_var else <=) min(df["column"])

Edit: can anyone explicitly show me how that logic would be implemented with the operator module?

while operator.gt(a, b) if bool_var else operator.eq(a, b): ?

Answer 1

To avoid duplicated blocks in your if/else branching all you need is to change your while statement basing on positive flag and comparison operator created by operator module:

from operator import eq, gt

def check_trends(df, set_lenght, positive=True, get_index=False):
    periods = []
    op = gt if positive else eq  # select operator
    
    for i in range(len(df.index)):
        period = 0
        while op(df["perc_btc"][i + period], min(df["perc_btc"])):
            if len(df.index) <= (i + period + 1):
                break
            period += 1
        if period > set_lenght:
            periods.append([i, period])
            if get_index:
                return [i, i + period]  # returns the last starting point
    return periods

Answer 2

Another way would be to make two separate comparisons, each contingent on newbool . In this example a is df["column"][i + period] and b is min(df["column"] .

>>> newbool = True
>>> a = 4
>>> b = 5
>>> (a>b and newbool) or (a<=b and not newbool)
False
>>> newbool = False
>>> (a>b and newbool) or (a<=b and not newbool)
True
>>> a = 6
>>> (a>b and newbool) or (a<=b and not newbool)
False
>>> newbool = True
>>> (a>b and newbool) or (a<=b and not newbool)
True
>>>

---

df["column"][i + period]

Can probably be written as

df.loc[i+period,"column"]

If it can it should be. Different choices for indexing .

---

To prevent a very long while statement the terms could be assigned to names before and at the bottom of the loop.

a = df.loc[i+period,"column"]>min(df["column"]
b = min(df["column"]
while (a>b and newbool) or (a<=b and not newbool)):
    # all the other stuff
    ...
    a = df.loc[i+period,"column"]
    b = min(df["column"])