How many extra keypresses on an old-style cell phone keyboard?

keypresses.py

# How many extra keypresses does it take to type alphabetic characters on an
# old-style cell phone keyboard?  
# https://www.facebook.com/zorikh.lequidre/posts/10151583168686662

# The result: 1.14908 extra keypresses "wasted" each time!

# Frequency table from here:  http://en.wikipedia.org/wiki/Letter_frequency
frequencies = {
 'a': 0.08167,
 'b': 0.01492,
 'c': 0.02782,
 'd': 0.04253,
 'e': 0.12702,
 'f': 0.02228,
 'g': 0.02015,
 'h': 0.06094,
 'i': 0.06966,
 'j': 0.00153,
 'k': 0.00772,
 'l': 0.04025,
 'm': 0.02406,
 'n': 0.06749,
 'o': 0.07507,
 'p': 0.01929,
 'q': 0.00095,
 'r': 0.05987,
 's': 0.06327,
 't': 0.09056,
 'u': 0.02758,
 'v': 0.00978,
 'w': 0.02360,
 'x': 0.00150,
 'y': 0.01974,
 'z': 0.00074,
}

keypresses = {
 'a': 1,
 'b': 2,
 'c': 3,
 'd': 1,
 'e': 2,
 'f': 3,
 'g': 1,
 'h': 2,
 'i': 3,
 'j': 1,
 'k': 2,
 'l': 3,
 'm': 1,
 'n': 2,
 'o': 3,
 'p': 1,
 'q': 2,
 'r': 3,
 's': 4,
 't': 1,
 'u': 2,
 'v': 3,
 'w': 1,
 'x': 2,
 'y': 3,
 'z': 4,
}

keypresses = sum(frequencies[key] * presses for (key, presses) in keypresses.items())
print 'Extra keypresses = %s' % (keypresses - 1)