change vars in place w/loop or list comprehension

J

Jatinder Singh

Hi
what u can do over here is
add a,b,c... in a list e.g. list.append(vars..)
and then use the statement
newlist = map(lambda x:x.upper(),list)
Now ur newlist will contain the modified list.
HOPE THIS THE BETTER SOLUTION TO UR PROBLEM


Quoting (e-mail address removed):
I am a python newbie, and am grappling with a fundamental concept. I
want to
modify a bunch of variables in place. Consider the following:
a = 'one'
b = 'two'
c = 'three'
list = [a, b, c]
for i in range(len(list)):
... list = list.upper()
...
'ONE'

or, better:
a = 'one'
b = 'two'
c = 'three'
[a, b, c] = [s.upper() for s in [a, b, c]]
a
'ONE'

Both of these accomplish what I'm after; I prefer the second for its
brevity.
But either approach requires that I spell out my list of vars-to-alter
[a, b, c] twice. This strikes me as awkward, and would be difficult
with longer lists. Any suggestions for a better way to do this?

--Ken
 
K

ken.boss

I am a python newbie, and am grappling with a fundamental concept. I
want to
modify a bunch of variables in place. Consider the following:
a = 'one'
b = 'two'
c = 'three'
list = [a, b, c]
for i in range(len(list)):
.... list = list.upper()
....
'ONE'

or, better:
a = 'one'
b = 'two'
c = 'three'
[a, b, c] = [s.upper() for s in [a, b, c]]
a
'ONE'

Both of these accomplish what I'm after; I prefer the second for its
brevity.
But either approach requires that I spell out my list of vars-to-alter
[a, b, c] twice. This strikes me as awkward, and would be difficult
with longer lists. Any suggestions for a better way to do this?

--Ken
 
R

Raymond Hettinger

a = 'one'
b = 'two'
c = 'three'
[a, b, c] = [s.upper() for s in [a, b, c]]
a
'ONE'

Both of these accomplish what I'm after; I prefer the second for its
brevity. But either approach requires that I spell out my list of
vars-to-alter [a, b, c] twice. This strikes me as awkward, and
would be difficult
with longer lists.

seq = a, b, c
a, b, c = seq = [s.upper() for s in seq]
 
P

Peter Hansen

I am a python newbie, and am grappling with a fundamental concept. I
want to
modify a bunch of variables in place. Consider the following:
[snip]
a = 'one'
b = 'two'
c = 'three'
[a, b, c] = [s.upper() for s in [a, b, c]]

Both of these accomplish what I'm after; I prefer the second for its
brevity.
But either approach requires that I spell out my list of vars-to-alter
[a, b, c] twice. This strikes me as awkward, and would be difficult
with longer lists. Any suggestions for a better way to do this?

While the real question is why you think you need to modify a bunch of
variables in place**, here's one solution to the immediate problem:

for name in 'a b c'.split():
globals()[name] = globals()[name].upper()

That, of course, will work only if they really are globals, and not if
they are local variables inside a function. If that's the case, you
really want to rethink this anyway (or use Raymond's approach, though it
still appears to require spelling out your list of names twice).

** Why you might not want to do this at all:

It's very rare in real code to need to work with a large number of
variables as a set like that. More generally you should probably be
storing the strings in something like a list and manipulating them
there. This would look like this:

shtuff = ['a', 'b', 'c']
for i,item in enumerate(shtuff):
shtuff = item.upper()

Or, even more likely:

newList = []
for item in shtuff:
newList.append(item.upper())

shtuff = newList


If you think you really need to do this anyway, consider posting an
example of real code from the program you are trying to write,
explaining the actual use case, and we can analyze the real situation
instead of debating contrived examples.

-Peter
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Members online

No members online now.

Forum statistics

Threads
473,744
Messages
2,569,484
Members
44,903
Latest member
orderPeak8CBDGummies

Latest Threads

Top