os.listdir

[Duncan Booth]

"Terry Reedy" <tjreedy at udel.edu> wrote in news:wSmdnYp3_9zFSf2iU-
KYvA at comcast.com:

> 
> "Duncan Booth" <duncan at NOSPAMrcp.co.uk> wrote in message
>> In C Python, all dictionaries have an initial size of 8 elements.
> There is
>> no way to override this (except recompiling!)
> 
> In you initialize a dict from a list of, say, 1000 (key,value) pairs,
> does it really disregard list size?  at least after first 2/3rds
> fillup?
> 
Yes. Very roughly speaking dict.__init__ looks like this (except in C):

 def __init__(self, dict, **kw):
    	if hasattr(dict, 'keys'):
    	    	self.update(dict)
     else:
    	    	for item in iter(dict):
               fast = tuple_or_list(item)
               assert len(fast)==2
               self[fast[0]] = fast[1]

The important point being that the code accepts an iterator rather than a 
list, so it doesn't know how many items will be returned. Of course there 
is nothing to stop someone adding a check whether the argument is really a 
list, and if so preallocating the size, but someone would have to evaluate 
whether the increased complexity, and longer runtime for smaller 
dictionaries justified the gain.

There's another twist to this:

   d = dict(aList)

At some point you have to create 'aList', and that too will start off small 
and double in size as it grows.

Even if you just write a constant dictionary:

   d = { 1:1, 2:2 }

the bytecode for this creates an empty dictionary, then evaluates each key, 
value pair and stores it in the dictionary.

-- 
Duncan Booth                                             duncan at rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?