accumulate -- apply a binary operator repeatedly

Suppose L=\{x0, x1, ..., xn\}. Then for any binary operator f, accumulate(f, L) returns the list \{f(x0, x1), f(f(x0, x1), x2), ...\} . In other words, the binary operator is applied to the first two elements of L, then to that result along with the next unused element of L, and so forth.

If accumulate(f, x, L) is called, the element x is used as the first argument of the binary function f. In other words, accumulate(f, \{x0, x1, \ldots, xn\}) is equivalent to accumulate(f, x0, \{x1, \ldots, xn\}).

The function accumulate(\{x_0, x_1, \ldots, x_n\}, f) returns the list \{..., f(x_{n-2}, f(x_{n-1}, x_n)), f(x_{n-1}, x_n) \} . That is, f is applied to the last two elements of the list, and the result placed at the end of the output. Then the accumulation proceeds backwards through the list. The optional argument x in accumulate(L, x, f) is used as the second argument in the first evaluation of f. So accumulate(\{x_0, x_1, \ldots, x_{n-1}\}, x_n, f) is equivalent to accumulate(\{x_0, x_1, \ldots, x_n\}, f).

If L is an instance of class the the iterator method installed, e.g., a string, then it may also be used with accumulate, but only the versions with f as the first argument. Its return value in this case is an Iterator object.

The difference between fold and accumulate is that fold returns the final result of all the nested evaluations of f, while accumulate lists all the intermediate values as well.