python: getting around division by zero

The answer given by Enrico is nice, but both solutions result in a warning:

RuntimeWarning: divide by zero encountered in log

As an alternative, we can still use the where function but only execute the main computation where it is appropriate:

# alternative implementation -- a bit more typing but avoids warnings.
loc = np.where(myarray>0)
result2 = np.zeros_like(myarray, dtype=float)
result2[loc] =np.log(myarray[loc])

# answer from Enrico...
myarray= np.random.randint(10,size=10)
result = np.where(myarray>0, np.log(myarray), 0)

# check it is giving right solution:
print(np.allclose(result, result2))

My use case was for division, but the principle is clearly the same:

x = np.random.randint(10, size=10)
divisor = np.ones(10,)
divisor[3] = 0 # make one divisor invalid

y = np.zeros_like(divisor, dtype=float)
loc = np.where(divisor>0) # (or !=0 if your data could have -ve values)
y[loc] = x[loc] / divisor[loc]

You are using a np function, so I can safely guess that you are working on a numpy array? Then the most efficient way to do this is to use the where function instead of a for loop

myarray= np.random.randint(10,size=10)
result = np.where(myarray>0, np.log(myarray), 0)

otherwise you can simply use the log function and then patch the hole:

myarray= np.random.randint(10,size=10)
result = np.log(myarray)
result[result==-np.inf]=0

The np.log function return correctly -inf when used on a value of 0, so are you sure that you want to return a 0? if somewhere you have to revert to the original value, you are going to experience some problem, changing zeros into ones...

You can do this.

def safe_ln(x):
   try:
      l = np.log(x)
   except ZeroDivisionError:
      l = 0
   return l

Since the log for x=0 is minus infinite, I'd simply check if the input value is zero and return whatever you want there:

def safe_ln(x):
    if x <= 0:
        return 0
    return math.log(x)

EDIT: small edit: you should check for all values smaller than or equal to 0.

EDIT 2: np.log is of course a function to calculate on a numpy array, for single values you should use math.log. This is how the above function looks with numpy:

def safe_ln(x, minval=0.0000000001):
    return np.log(x.clip(min=minval))