Hi guys I would like to loop multiple elements in a list elements creating a for loop, to make this script work with multi selection.. at the moment I can find an appropriate solution to append each element contained in myStr... Any idea? (at the moment I've added myStr[0] to pick first element)
myStr = cmds.ls(sl=1)
for i in myStr:
splits = myStr[0].split('_')
ver_up = int(splits[-2]) + 1
splits[-2] = '%04d'%ver_up
newStr = '_'.join(splits)
print(newStr)
cmds.duplicate(n=newStr)
If I understand you correctly, then you have several objects with a name based on <objectName>_<version>_<otherIdentifiers>. And you want to duplicate them and count up the version number. Correct? If so, you just have to add the object you want to duplicate to the cmds.duplicate() command like this:
cmds.duplicate(i, n=newStr)
Related
I'm trying to swap str elements in list with same elements but capitalize their first letter.
While trying to achieve this I'm just step by stepping it and when I try to use for loop to just append list with capitalized elements, my compiler freezes and proceeds to gradually increase in RAM usage up to 90%.
I can guess it has to do something with built in functions that I use (probably incorrectly). Can anyone help me understand what is happening and how should I approach it?
Here is code:
title = 'a clash of KINGS'
out = title.split()
for i in out:
out.append(i.capitalize())
Don't change a list while iterating over it. You keep adding elements to out list. You can print out inside the loop and see for yourself. Even if it didn't enter into infinite loop, still you did not replace the initial values, but just add more and more elements.
you can use list comprehension
title = 'a clash of KINGS'
out = title.split()
out = [word.capitalize() for word in out]
you can combine last 2 lines into one
title = 'a clash of KINGS'
out = [word.capitalize() for word in title.split()]
I think you are in a infinite loop. You're not accessing the out element, you're keep appending a lot of elements inside the list. I think what you're trying to do is:
title = 'a clash of KINGS'
out = title.split()
for i in range(len(out)):
out[i] = out[i].capitalize()
This seems like a fairly straightforward problem but I can't seem to find an efficient way to do it. I have a list of lists like this:
list = [['abc','def','123'],['abc','xyz','123'],['ghi','jqk','456']]
I want to get a list of unique entries by the third item in each child list (the 'id'), i.e. the end result should be
unique_entries = [['abc','def','123'],['ghi','jqk','456']]
What is the most efficient way to do this? I know I can use set to get the unique ids, and then loop through the whole list again. However, there are more than 2 million entries in my list and this is taking too long. Appreciate any pointers you can offer! Thanks.
How about this: Create a set that keeps track of ids already seen, and only append sublists where id's where not seen.
l = [['abc','def','123'],['abc','xyz','123'],['ghi','jqk','456']]
seen = set()
new_list = []
for sl in l:
if sl[2] not in seen:
new_list.append(sl)
seen.add(sl[2])
print new_list
Result:
[['abc', 'def', '123'], ['ghi', 'jqk', '456']]
One approach would be to create an inner loop. within the first loop you iterate over the outer list starting from 1, previously you will need to create an arraylist which will add the first element, inside the inner loop starting from index 0 you will check only if the third element is located as part of the third element within the arraylist current holding elements, if it is not found then on another arraylist whose scope is outside outher loop you will add this element, else you will use "continue" keyword. Finally you will print out the last arraylist created.
Im pretty new to Python.
I have a list which looks like the following:
list = [('foo,bar,bash',)]
I grabbed it from and sql table (someone created the most rubbish sql table!), and I cant adjust it. This is literally the only format I can pull it in. I need to chop it up. I can't split it by index:
print list[0]
because that just literally gives me:
[('foo,bar,bash',)]
How can I split this up? I want to split it up and write it into another list.
Thank you.
list = [('foo,bar,bash',)] is a list which contains a tuple with 1 element. You should also use a different variable name instead of list because list is a python built in.
You can split that one element using split:
lst = [('foo,bar,bash',)]
print lst[0][0].split(',')
Output:
['foo', 'bar', 'bash']
If the tuple contains more than one element, you can loop through it:
lst = [('foo,bar,bash','1,2,3')]
for i in lst[0]:
print i.split(',')
I have a list of URLs in an open CSV which I have ordered alphabetically, and now I would like to iterate through the list and check for duplicate URLs. In a second step, the duplicate should then be removed from the list, but I am currently stuck on the checking part which I have tried to solve with a nested for-loop as follows:
for i in short_urls:
first_url = i
for s in short_urls:
second_url = s
if i == s:
print "duplicate"
else:
print "all good"
The print statements will obviously be replaced once the nested for-loop is working. Currently, the list contains a few duplicates, but my nested loop does not seem to work correctly as it does not recognise any of the duplicates.
My question is: are there better ways to do perform this exercise, and what is the problem with the current nested for-loop?
Many thanks :)
By construction, your method is faulty, even if you indent the if/else block correctly. For instance, imagine if you had [1, 2, 3] as short_urls for the sake of argument. The outer for loop will pick out 1 to compare to the list against. It will think it's finding a duplicate when in the inner for loop it encounters the first element, a 1 as well. Essentially, every element will be tagged as a duplicate and if you plan on removing duplicates, you'll end up with an empty list.
The better solution is to call set(short_urls) to get a set of your urls with the duplicates removed. If you want a list (as opposed to a set) of urls with the duplicates removed, you can convert the set back into a list with list(set(short_urls)).
In other words:
short_urls = ['google.com', 'twitter.com', 'google.com']
duplicates_removed_list = list(set(short_urls))
print duplicates_removed_list # Prints ['google.com', 'twitter.com']
if i == s:
is not inside the second for loop. You missed an indentation
for i in short_urls:
first_url = i
for s in short_urls:
second_url = s
if i == s:
print "duplicate"
else:
print "all good"
EDIT: Also you are comparing every element of an array with every element of the same array. This means compare the element at position 0 with the element at postion 0, which is obviously the same.
What you need to do is starting the second for at the position after that reached in the first for.
I'm coming up with a rather trivial problem, but since I'm quite new to python, I'm smashing my head to my desk for a while. (Hurts). Though I believe that's more a logical thing to solve...
First I have to say that I'm using the Python SDK for Cinema 4D so I had to change the following code a bit. But here is what I was trying to do and struggling with:
I'm trying to group some polygon selections, which are dynamically generated (based on some rules, not that important).
Here's how it works the mathematical way:
Those selections are based on islands (means, that there are several polygons connected).
Then, those selections have to be grouped and put into a list that I can work with.
Any polygon has its own index, so this one should be rather simple, but like I said before, I'm quite struggling there.
The main problem is easy to explain: I'm trying to access a non existent index in the first loop, resulting in an index out of range error. I tried evaluating the validity first, but no luck. For those who are familiar with Cinema 4D + Python, I will provide some of the original code if anybody wants that. So far, so bad. Here's the simplified and adapted code.
edit: Forgot to mention that the check which causes the error actually should only check for duplicates, so the current selected number will be skipped since it hal already been processed. This is necessary due to computing-heavy calculations.
Really hope, anybody can bump me in the right direction and this code makes sense so far. :)
def myFunc():
sel = [0,1,5,12] # changes with every call of "myFunc", for example to [2,8,4,10,9,1], etc. - list alway differs in count of elements, can even be empty, groups are beeing built from these values
all = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] # the whole set
groups = [] # list to store indices-lists into
indices = [] # list to store selected indices
count = 0 # number of groups
tmp = [] # temporary list to copy the indices list into before resetting
for i in range(len(all)): # loop through values
if i not in groups[count]: # that's the problematic one; this one actually should check whether "i" is already inside of any list inside the group list, error is simply that I'm trying to check a non existent value
for index, selected in enumerate(sel): # loop through "sel" and return actual indices. "selected" determines, if "index" is selected. boolean.
if not selected: continue # pretty much self-explanatory
indices.append(index) # push selected indices to the list
tmp = indices[:] # clone list
groups.append(tmp) # push the previous generated list to another list to store groups into
indices = [] # empty/reset indices-list
count += 1 # increment count
print groups # debug
myFunc()
edit:
After adding a second list which will be filled by extend, not append that acts as counter, everything worked as expected! The list will be a basic list, pretty simple ;)
groups[count]
When you first call this, groups is an empty list and count is 0. You can't access the thing at spot 0 in groups, because there is nothing there!
Try making
groups = [] to groups = [[]] (i.e. instead of an empty list, a list of lists that only has an empty list).
I'm not sure why you'd want to add the empty list to groups. Perhaps this is better
if i not in groups[count]:
to
if not groups or i not in groups[count]:
You also don't need to copy the list if you're not going to use it for anything else. So you can replace
tmp = indices[:] # clone list
groups.append(tmp) # push the previous generated list to another list to store groups into
indices = [] # empty/reset indices-list
with
groups.append(indices) # push the previous generated list to another list to store groups into
indices = [] # empty/reset indices-list
You may even be able to drop count altogether (you can always use len(groups)). You can also replace the inner loop with a listcomprehension
def myFunc():
sel = [0,1,5,12] # changes with every call of "myFunc", for example to [2,8,4,10,9,1], etc. - list alway differs in count of elements, can even be empty, groups are beeing built from these values
all = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] # the whole set
groups = [] # list to store indices-lists into
for i in range(len(all)): # loop through values
if not groups or i not in groups[-1]: # look in the latest group
indices = [idx for idx, selected in enumerate(sel) if selected]
groups.append(indices) # push the previous generated list to another list to store groups into
print groups # debug
correct line 11 from:
if i not in groups[count]
to:
if i not in groups: