Better looping, for string manipulation (python)

Question

If i have this code

s = 'abcdefghi'
for grp in (s[:3],s[3:6],s[6:]):
    print "'%s'"%(grp)

    total = calc_total(grp)

    if (grp==s[:3]):
        # more code than this
        p = total + random_value
        x1 = my_function(p)

    if (grp==s[3:6]):
        # more code than this
        p = total + x1
        x2 = my_function(p)

    if (grp==s[6:]):
        # more code than this
        p = total + x2
        x3 = my_function(p)

If the group is the first group, perform code for this group, if the group is the second group, perform code using the a value generated from code performed for the first group, the same applies for the third group, using a generated value from code for the second group:

How can i tidy this up to use better looping?

Thanks

Peter Westlake · Accepted Answer · 2010-03-02 18:13:35Z

3

I may have misunderstood what you're doing, but it appears that you want to do something to s[:3] on the first iteration, something different to s[3:6] on the second, and something else again to s[6:] on the third. In other words, that isn't a loop at all! Just write those three blocks of code out one after another, with s[:3] and so on in place of grp.

answered Mar 2, 2010 at 18:13

Peter Westlake

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1 Comment

joec Over a year ago

Sorry, i forgot to include my total = calc_total(grp) statement

Erlend Kristiansen · Accepted Answer · 2012-04-11 11:50:40Z

1

I must say I agree with Peter in that the loop is redundant. If you are afraid of duplicating code, then just move the repeating code into a function and call it multiple times:

s = 'abcdefghi'

def foo(grp):
    # Anything more you would like to happen over and over again
    print "'%s'"%(grp)
    return calc_total(grp)

def bar(grp, value):
    total = foo(grp)
    # more code than this
    return my_function(total + value)

x1 = bar(s[:3], random_value)
x2 = bar(s[3:6], x1)
x3 = bar(s[6:], x2)

If

    # more code than this

contains non-duplicate code, then you must of course move that out of "bar" (which together with "foo" should be given a more descriptive name).

answered Apr 11, 2012 at 11:50

Erlend Kristiansen

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Comments

Alex Martelli · Accepted Answer · 2010-03-02 18:52:09Z

0

I'd code something like this as follows:

for i, grp in enumerate((s[:3],s[3:6],s[6:])):
    print "'%s'"%(grp)

    total = calc_total(grp)
    # more code that needs to happen every time

    if i == 0:
        # code that needs to happen only the first time
    elif i == 1:
        # code that needs to happen only the second time

etc. The == checks can be misleading if one of the groups "just happens" to be the same as another one, while the enumerate approach runs no such risk.

answered Mar 2, 2010 at 18:52

Alex Martelli

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Comments

ephemient · Accepted Answer · 2010-03-02 18:45:23Z

0

x = reduce(lambda x, grp: my_function(calc_total(list(grp)) + x),
        map(None, *[iter(s)] * 3), random_value)

At the end, you'll have the last x.

Or, if you want to keep the intermediary results around,

x = []
for grp in map(None, *[iter(s)] * 3):
    x.append(my_function(calc_total(list(grp)) + (x or [random_value])[-1]))

Then you have x[0], x[1], x[2].

edited Mar 2, 2010 at 18:45

answered Mar 2, 2010 at 18:40

ephemient

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Comments

jcdyer · Accepted Answer · 2010-03-02 19:24:30Z

0

Get your data into the list you want, then try the following:

output = 0
seed = get_random_number()
for group in input_list:
    total = get_total(group)
    p = total + seed
    seed = my_function(p)

input_list will need to look like ['abc', 'def', 'ghi']. But if you want to extend it to ['abc','def','ghi','jkl','mno','pqr'], this should still work.

answered Mar 2, 2010 at 19:24

jcdyer

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