# Python For Loops, using range vs enumerate

Python for loops are different to the conventional programming languages in a key aspect, its syntax doesn't use an iterator.

But that is a feature not a bug, the '*Pythonic*' way to write for loop is to run it over a range or a list by item and not by using an index to refer to the next element like C or Java. E.g.

```
fruits = ['apple', 'orange', 'banana', 'tomato', 'cucumber']
for fruit in fruits:
print(fruit)
```

Will print out

```
apple
orange
banana
tomato
cucumber
```

If we wanted to do the same in C, not withstanding the complexities, it would look like,

```
for (int i = 0, i < 5, i++)
{
printf("%s", fruits[i])
}
```

## Range and Enumerate functions

But there are many scenarios where you might need to iterate using index. For such cases, Python has two in-builts functions `range()`

and `enumerate()`

that provides this feature.

### range()

Range is used to iterate over a sequence of numbers, e.g. to print 0 - 4,

```
for i in range(5):
print(i)
```

We can use this to iterate over out fruits list with an iterator

```
fruits = ['apple', 'orange', 'banana', 'tomato', 'cucumber']
for i in range(5):
print(fruits[i])
```

But you now have a new problem, in the first example, we didn't have to worry about the length of the list, with range you do. You can still solve it by usng `len()`

method

```
fruits = ['apple', 'orange', 'banana', 'tomato', 'cucumber']
for i in range(len(fruits)):
print(fruits[i])
```

### enumerate()

Instead of calculating the length and iterating over the list, we can also use enumerate() to get the same results.

```
fruits = ['apple', 'orange', 'banana', 'tomato', 'cucumber']
for i, item in enumerate(fruits):
print(f"Using iterator: {fruits[i]}")
print(f"Using item: {item}")
```

## Range vs Enumerate: What should you use?

As with most things, the answer is, it depends! And more often than not, it will end up being a personal choice.

But from a performance perspective, we can test it.

### Performance testing range() and enumerate()

Let's start by setting up a baseline. We will generate a list with 10000 integers, and then compute if each one of them are prime number of not and add them to an output list.

#### Baseline: Simple iteration

```
from time import time
# start the time counter
start = time()
def prime(num):
'''Function to compute if a number is prime'''
for i in range(2, int(num/2)+1):
if num % i == 0:
return False
return True
# Generating a list of integers
n = 9999
inputs = [i for i in range(n)]
outputs = []
for number in inputs:
outputs.append(f"{number} is prime? {prime(number)}")
end = time()
print(end - start)
```

For me it took ~44.6 secs

#### Using range()

```
from time import time
# start the time counter
start = time()
def prime(num):
'''Function to compute if a number is prime'''
for i in range(2, int(num/2)+1):
if num % i == 0:
return False
return True
# Generating a list of integers
n = 9999
inputs = [i for i in range(n)]
outputs = []
for i in range(len(inputs)):
outputs.append(f"{inputs[i]} is prime? {prime(inputs[i])}")
end = time()
print(end - start)
```

Took roughly ~47.7 seconds

### Using enumerate()

```
from time import time
# start the time counter
start = time()
def prime(num):
'''Function to compute if a number is prime'''
for i in range(2, int(num/2)+1):
if num % i == 0:
return False
return True
# Generating a list of integers
n = 9999
inputs = [i for i in range(n)]
outputs = []
for i, num in enumerate(inputs):
outputs.append(f"{i} is prime? {prime(num)}")
end = time()
print(end - start)
```

This took about ~45.76 seconds

## Conclusion

If you're iterating over a list, enumerate is probably the most optimal option.