# Sum a list of numbers in Python

## Sum a list of numbers in Python

### Question

I have a list of numbers such as `[1,2,3,4,5...]`

, and I want to calculate `(1+2)/2`

and for the second, `(2+3)/2`

and the third,
`(3+4)/2`

, and so on. How can I do that?

I would like to sum the first number with the second and divide it by 2, then sum the second with the third and divide by 2, and so on.

Also, how can I sum a list of numbers?

```
a = [1, 2, 3, 4, 5, ...]
```

Is it:

```
b = sum(a)
print b
```

to get one number?

This doesn't work for me.

### Popular Answer

Question 1: So you want (element 0 + element 1) / 2, (element 1 + element 2) / 2, ... etc.

We make two lists: one of every element except the first, and one of every element except the last. Then the averages we want are the averages of each pair taken from the two lists. We use `zip`

to take pairs from two lists.

I assume you want to see decimals in the result, even though your input values are integers. By default, Python does integer division: it discards the remainder. To divide things through all the way, we need to use floating-point numbers. Fortunately, dividing an int by a float will produce a float, so we just use `2.0`

for our divisor instead of `2`

.

Thus:

```
averages = [(x + y) / 2.0 for (x, y) in zip(my_list[:-1], my_list[1:])]
```

Question 2:

That use of `sum`

should work fine. The following works:

```
a = range(10)
# [0,1,2,3,4,5,6,7,8,9]
b = sum(a)
print b
# Prints 45
```

Also, you don't need to assign everything to a variable at every step along the way. `print sum(a)`

works just fine.

You will have to be more specific about exactly what you wrote and how it isn't working.

Read more... Read less...

Sum list of numbers:

```
sum(list_of_nums)
```

Calculating half of n and n - 1 (if I have the pattern correct), using a list comprehension:

```
[(x + (x - 1)) / 2 for x in list_of_nums]
```

Sum adjacent elements, e.g. ((1 + 2) / 2) + ((2 + 3) / 2) + ... using reduce and lambdas

```
reduce(lambda x, y: (x + y) / 2, list_of_nums)
```

**Question 2:**
To sum a list of integers:

```
a = [2, 3, 5, 8]
sum(a)
# 18
# or you can do:
sum(i for i in a)
# 18
```

If the list contains integers as strings:

```
a = ['5', '6']
# import Decimal: from decimal import Decimal
sum(Decimal(i) for i in a)
```

You can try this way:

```
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
sm = sum(a[0:len(a)]) # Sum of 'a' from 0 index to 9 index. sum(a) == sum(a[0:len(a)]
print(sm) # Python 3
print sm # Python 2
```

```
>>> a = range(10)
>>> sum(a)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not callable
>>> del sum
>>> sum(a)
45
```

It seems that `sum`

has been defined in the code somewhere and overwrites the default function. So I deleted it and the problem was solved.

Using a simple `list-comprehension`

and the `sum`

:

```
>> sum(i for i in range(x))/2. #if x = 10 the result will be 22.5
```