Create if-elif statements using for loop

Question

I am dealing with an issue that involves multiple if and elif conditining..precisely stating, my case goes as follows:

if len(g) == 2:
   a = 'rea: 300'
   b = 'ref: "%s": {"sds": 200},"%s": {"sds": 300}' % (g[0],g[1])

elif len(g) == 3:
   a = 'rea: 400'
   b = 'ref: "%s": {"sds": 200},"%s": {"sds": 300},"%s": {"sds": 400}' % (g[0],g[1],g[2])
....

And this elif conditioning is supposed to go up to elif len(g) == 99...so I suppose there should be some elegant way to do this. Moreover, if you observe, there is a pattern with which the 'rea' and 'ref' are progressing, which can be stated as:

 if len(g) == x:
    a = 'rea: (x*100)+100'
    b = 'ref: "%s": {"sds": 200},"%s": {"sds": 300},"%s": {"sds": (x*100)+100}' % (g[0],g[1],g[2])

is b supposed to be a dictionary? Why recreate it as as string? — Inbar Rose
– Inbar Rose, Commented Jun 30, 2013 at 7:31

Burhan Khalid · Accepted Answer · 2013-06-30 07:21:22Z

2

Maybe something like this:

g_len = len(g)
a = "rea: {}".format((g_len + 1) * 100)
b = "ref: "
for i, g_i in enumerate(g):
    b += ' "{}": {{"sds": {}}},'.format(g_i, (i+2) * 100)

edited Jun 30, 2013 at 7:21

Burhan Khalid

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answered Jun 30, 2013 at 6:45

mishik

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5 Comments

user2480542 Over a year ago

This is by far the most elegant one..thanks Mishik and Burhan! :-)

user2480542 Over a year ago

one more thing guys..how to implement this on Python 2.6 or 2.4? There maybe a little syntax change i believe..

mishik Over a year ago

I don't really remember much of 2.6 issues. What's not working?

user2480542 Over a year ago

IT is raising this exception: b += ' "{}": {{"sds": {}}},'.format(g_i, (i+2) * 100) ValueError: zero length field name in format

mishik Over a year ago

Hm... I guess it does not like {}. Try using "{0}": {{"sds": {1}}},

Inbar Rose · Accepted Answer · 2013-06-30 07:37:53Z

2

Try this method:

def func(g):
    if not 1 < len(g) < 100:
        raise ValueError('inadequate length')
    d = {x:{'sds':(i+2)*100} for i, x in enumerate(g)}
    a = 'rea: %s00' % (len(g)+1)
    b = 'ref: %s' % str(d)[1:-1]
    return (a, b)

I don't know why you are creating a string b which looks very much like a dictionary, but I am sure you have your reasons...

>>> func(range(3))
('rea: 400', "ref: 0: {'sds': 200}, 1: {'sds': 300}, 2: {'sds': 400}")

answered Jun 30, 2013 at 7:37

Inbar Rose

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Comments

TerryA · Accepted Answer · 2013-06-30 08:07:53Z

0

Use a dictionary:

x = 100
d = {}
for i in xrange(2, len(g)+1):
    d[i] = ['rea: {}'.format(100+x*i), 'ref: '+ ('%s: {"sds": 200}, ' *(i-1) + ' %s: {"sds": 200}') % tuple(g[:i])]

Now, d will look something like:

{2: ['rea: 300',
     'ref: "depends_on_g": {"sds": 200}, "depends_on_g": {"sds": 300}'],
 3: ['rea: 400',
     'ref: "depends_on_g": {"sds": 200}, "depends_on_g": {"sds": 300}', "depends_on_g": {"sds": 300}]
 ...
}

Then, to access it:

a, b = d.get(len(g))

No if-statements needed :)

edited Jun 30, 2013 at 8:07

answered Jun 30, 2013 at 6:49

TerryA

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5 Comments

mishik Over a year ago

"ref" value should be increasing for each i

TerryA Over a year ago

@mishik I just notcied that :)

mishik Over a year ago

And you now have a dict of n^2 size, storing all the non required values :)

TerryA Over a year ago

@mishik What :p? Firstly, my dict is length 98, and secondly, it gets the contents from g fine.

mishik Over a year ago

The dict is 98 length alright, but d[1] is X long, d[1] is 2X long, etc... d[99] ix 99X long :) X + 2X + ... + 99X = 4950X

Steve Barnes · Accepted Answer · 2013-06-30 07:16:49Z

-1

Personally I would go for something functional at the time that the length was known like:

def do_Condition(g):
    """ Condtion the result based on the length of g """
    l = len(g)
    a = 'rea: %d' % (100 + 100*l)
    items = ['ref: "%s"' % (g[0])]
    n = 100
    for i in g[1:]:
        n += 100
        items.append('{"sds": %d},"%s"' % (n, i))
    b = ': '.join(items)
    return (a, b)

answered Jun 30, 2013 at 7:16

Steve Barnes

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1 Comment

Inbar Rose Over a year ago

You don't need to store the len(g) as a variable, the length is always calculated and calling the length method on a container does not re-evaluate each time, it only retrieves the value, it is actually slower to store length as a separate value.

rtrwalker · Accepted Answer · 2013-06-30 08:41:15Z

-1

this seems to work:

a = 'rea: %d00' % (len(g)+1)
b = ",".join(['ref: "%s": {"sds": %d00}' % (i, j) for i, j in  zip(g, range(2,len(g)+2))])

answered Jun 30, 2013 at 8:41

rtrwalker

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