I've got the following code:
def chunk_trades(A): last = A[0] new = [] for x in A.iteritems(): if np.abs((x[1]-last)/last) > 0.1: new.append(x[1]) last = x[1] else: new.append(last) s = pd.Series(new, index=A.index) return s Sometimes last can be zero. In this case, I'd like it to just carry on gracefully as if last was almost zero.
What's the cleanest way?
Just Replace your line by this:
if not last or np.abs((x[1]-last)/last) > 0.1: This will not raise an exception since the left assertion is checked first.
Not sure if you would really want to divide by "almost 0", since the result will be "almost infinity", but you can also do this:
if last == 0: last = sys.float_info.min This is the minimum positive normalized float, i.e. the value closest to zero.
Source: https://docs.python.org/2/library/sys.html#sys.float_info
If I anderstand correctly, when last == 0 youl'll get ZeroDivisionError, won't you? If yes, please consider following slightly modified version of your code:
def chunk_trades(A): last = A[0] new = [] for x in A.iteritems(): try: if np.abs((x[1]-last)/last) > 0.1: new.append(x[1]) last = x[1] else: new.append(last) except ZeroDivisionError: eps = 1e-18 # arbitary infinitesmall number last = last + eps if np.abs((x[1]-last)/last) > 0.1: new.append(x[1]) last = x[1] else: new.append(last) s = pd.Series(new, index=A.index) return s 
1e-18
lastis zero, you want theifblock to execute, not theelseblock, right?if last == 0 or np.abs(...) > 0.1? Alternatively do exactly what you described, define anepsilon = 0.0000001and then do/ (last or epsilon). whenlast == 0it is considered false andepsilonwill be used in its place.