append to items depending on prior item

[M. Clift]

Hi All,

I have tried to come up with a way to do this myself and all I end up with
is very long code.

What I have is a say item1, item4, item2, item1 etc...

What I want to do is append to each item an extra value depending on the
previous item.

from random import *

items = [' item1',' item4',' item2',' item1']

items[0].append choice('1','2','3')
print items

for idx in range(len(items)):
if previous item == ['item1']:
next item.append choice('a','b','c')
if previous item == ['item2']:
next item.append choice('d','e','f')

print items
appended items = item1b, item4a, item2f etc...

Now I know that the code is hopelessly wrong - you can't append a string,
but you get the idea. Doing it using if statements could go on and on. Could
someone show me a short way

Thanks,

M

[M. Clift]

Hi All,

Of course that should be

next Item + (choice('a,b,c'))

[Peter Otten]

M. Clift wrote:

Now I know that the code is hopelessly wrong
[...]
Could someone show me a short way

items = "item1 item4 item2 item1".split()
[i + random.choice(dict(item1="abc", item2="def").get(p, [""])) for (p,

i) in zip(items[-1:]+items[:-1], items)]
['item1c', 'item4c', 'item2', 'item1f']
Peter

[Alex Martelli]

M. Clift <no***@here.com> wrote:
...

What I have is a say item1, item4, item2, item1 etc...

What I want to do is append to each item an extra value depending on the
previous item.

from random import *

items = [' item1',' item4',' item2',' item1']

items[0].append choice('1','2','3')
print items

for idx in range(len(items)):
if previous item == ['item1']:
next item.append choice('a','b','c')
if previous item == ['item2']:
next item.append choice('d','e','f')

print items

appended items = item1b, item4a, item2f etc...

This latest example totally contrast with everything else you're saying.
How could 'b' get appended to the first 'item1', for example, when
you're trying to use a choice among '1', '2' and '3'?! I'm going to try
to guess what you mean (if you had been a bit more precise in your
example of desired input and output it WOULD have been far better, of
course)...:

def weird_appender(sequence):
choices = dict({None: '123'}, item1='abc', item2='def')
previous = None
for item in sequence:
yield item+random.choice(choices.get(previous, choices[None]))
previous = item

and if for some super=weird reason you want to trample the contents of
list items rather than just operating on a sequence-in / sequence-out
basis, items[:]=weird_appender(items) will serve.
Alex

[M. Clift]

Hi Peter,

Thankyou, it looks perfect. At the risk of sounding dumb, however, what do I
print to get your output?

items = "item1 item4 item2 item1".split()
[i + random.choice(dict(item1="abc", item2="def").get(p, [""]))
for (p,i) in zip(items[-1:]+items[:-1], items)]
print ???

Thanks,

M

[Peter Otten]

M. Clift wrote:

Thankyou, it looks perfect. At the risk of sounding dumb, however, what do
I print to get your output?

You would print all the stuff inside [], i. e.

result = [i + random.choice(dict(item1="abc", item2="def").get(p, [""]))
for (p, i) in zip(items[-1:]+items[:-1], items)]
print result

but: the code was *not* meant to be perfect - rather as close as you can get
to obfuscation in Python.
You should first describe exactly what you want in terms of input/output.
Stating a problem precisely is often the hardest and largest part of its
solution. After that try to write a script as _simple_ and _clear_ as you
can (in Python as opposed to your pseudocode).
If you then ask on c.l.py for help with a specific problem you encounter or
for possible improvements, you'll end up with an idiomatic and
understandable solution which is not necessarily the shortest.
In short: short != good.

More practially:
- a plain old for-loop (like in Alex Martelli's post) with a variable to
remember the previous item beats my zip() magic in efficiency and - more
importantly - in clarity.
- if you encounter a long list of if ... elif ... statements, this can often
be simplified into a dictionary lookup.

HTH ...really,
Peter

[M. Clift]

Hi Peter,

Thankyou. You have been really helpful, I appreciate it.

Malcolm