Numerical Python question

[David Mertz]

mcrider at bigfoot.com (2mc) wrote previously:
|I apologize for not making this clear.  If an array were to be viewed
|as a spreadsheet, then rows would be individual days and columns would
|be date, open price, high price, etc.

Aside from the speed increase, Numeric provides quite a few handy syntax
tricks.  For example, in a crude version of your problem, I first create
the "spreadsheet", and populate it with silly open/close/high/low
prices.

  >>> from Numeric import *
  >>> stock = zeros((5,20),Int)
  >>> stock[0,:] = range(20)  # number the days
  >>> stock[1,:] = [10]*20    # open price
  >>> stock[2,:] = [11]*20    # close price
  >>> stock[3,:] = [13]*20    # high price
  >>> stock[4,:] = [8]*20     # low price
  >>> print stock
  [[ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19]
   [10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10]
   [11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11]
   [13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13]
   [ 8  8  8  8  8  8  8  8  8  8  8  8  8  8  8  8  8  8  8  8]]

Of course, it is an odd stock that opens, closes, and peaks at the same
price on every day. :-)

Next, I can find a particular average high for a range:

  >>> avg_high_day5to15 = sum(stocks[3,5:15])/len(stocks[3,5:15])
  >>> avg_high_day5to15
  13

You might want to generalize it in a function:

  >>> def ten_day_avg_high(beg):
  ...     return sum(stocks[3,beg:beg+10])/len(stocks[3,beg:beg+10])

Which let's you calculate your running averages:

  >>> map(ten_day_avg_high, range(20))
  [13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13,
  13, 13]

This works even near the end, where there are fewer than ten days to
average--we divide by the length of the tail, not by '10' to make sure
of that.

You can work out deviations and other statistics in similar ways.

Yours, David...

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