Eliminate "extra" variable

[Tim Chase]

On 2013-12-06 11:37, Igor Korot wrote:
> def MyFunc(self, originalData):
>      data = {}
>      for i in xrange(0, len(originalData)):
>            dateStr, freq, source = originalData[i]
>            data[str(dateStr)]  = {source: freq}

this can be more cleanly/pythonically written as

  def my_func(self, original_data):
    for date, freq, source in original_data
      data[str(date)] = {source: freq}

or even just

    data = dict(
      (str(date), {source: freq})
      for date, freq, source in original_data
      )

You're calling it a "dateStr", which suggests that it's already a
string, so I'm not sure why you're str()'ing it.  So I'd either just
call it "date", or skip the str(date) bit if it's already a string.
That said, do you even need to convert it to a string (as
datetime.date objects can be used as keys in dictionaries)?

>     for i in xrange(0, len(dateStrs) - 1):
>           currDateStr = str(dateStrs[i])
>           nextDateStrs = str(dateStrs[i + 1])
> 
> It seems very strange that I need the dateStrs list just for the
> purpose of looping thru the dictionary keys.
> Can I get rid of the "dateStrs" variable?

Your code isn't actually using the data-dict at this point.  If you
were doing something with it, it might help to know what you want to
do.

Well, you can iterate over the original data, zipping them together:

  for (cur, _, _), (next, _, _) in zip(
      original_data[:-1],
      original_data[1:]
      ):
    do_something(cur, next)

If your purpose for the "data" dict is to merely look up stats from
the next one, the whole batch of your original code can be replaced
with:

  for (
        (cur_dt, cur_freq, cur_source),
        (next_dt, next_freq, next_source)
        ) in zip(original_data[:-1], original_data[1:]):
    # might need to do str(cur_dt) and str(next_dt) instead?
    do_things_with(cur_dt, cur_freq, cur_source,
      next_dt, next_freq, next_source)

That eliminates the dict *and* the extra variable name. :-)

-tkc