I would like to know if there is a built in function in python for the equivalent Haskell `scanl`, as `reduce` is the equivalent of `foldl`.

Something that does this:

```
Prelude> scanl (+) 0 [1 ..10]
[0,1,3,6,10,15,21,28,36,45,55]

```

The question is not about how to implement it, I already have 2 implementations, shown below (however, if you have a more elegant one please feel free to show it here).

First implementation: 

```
 # Inefficient, uses reduce multiple times
 def scanl(f, base, l):
   ls = [l[0:i] for i in range(1, len(l) + 1)]
   return [base] + [reduce(f, x, base) for x in ls]

  print scanl(operator.add, 0, range(1, 11))

```

Gives:

```
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]

```

Second implementation:

```
 # Efficient, using an accumulator
 def scanl2(f, base, l):
   res = [base]
   acc = base
   for x in l:
     acc = f(acc, x)
     res += [acc]
   return res

 print scanl2(operator.add, 0, range(1, 11))

```

Gives:

```
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]

```

Thank you :)