list of lists of dictionaries? - list

I need to create a structure, in my mind similar to an array of linked lists (where a python list = array and dictionary = linked list). I have a list called blocks, and this is something like what I am looking to make:
blocks[0] = {dictionary},{dictionary},{dictionary},...
blocks[1] = {dictionary},{dictionary},{dictionary},...
etc..
currently I build the blocks as such:
blocks = []
blocks.append[()]
blocks.append[()]
blocks.append[()]
blocks.append[()]
I know that must look ridiculous. I just cannot see in my head what that just made, which is part of my problem. I assign to a block from a different list of dictionary items. Here is a brief overview of how a single block is created...
hold = {}
hold['file']=file
hold['count']=count
hold['mass']=mass_lbs
mg1.append(hold)
##this append can happen several times to mg1
blocks[i].append(mg1[j])
##where i is an index for the block I want to append to, and j is the list index corresponding to whichever dictionary item of mg1 I want to grab.
The reason I want these four main indices in blocks is so that I have shorter code with just the one list instead of block1 block2 block3 block4, which would just make the code way longer than it is now.

Okay, going off of what was discussed in the comments, you're looking for a simple way to create a structure that is a list of four items where each item is a list of dictionaries, and all the dictionaries in one of those lists have the same keys but not necessarily the same values. However, if you know exactly what keys each dictionary will have and that never changes, then it might be worth it to consider making them classes that wrap dictionaries and have each of the four lists be a list of objects. This would be easier to keep in your head, and a bit more Pythonic in my opinion. You also gain the advantage of ensuring that the keys in the dictionary are static, plus you can define helper methods. And by emulating the methods of a container type, you can still use dictionary syntax.
class BlockA:
def __init__(self):
self.dictionary = {'file':None, 'count':None, 'mass':None }
def __len__(self):
return len(self.dictionary)
def __getitem__(self, key):
return self.dictionary[key]
def __setitem__(self, key, value):
if key in self.dictionary:
self.dictionary[key] = value
else:
raise KeyError
def __repr__(self):
return str(self.dictionary)
block1 = BlockA()
block1['file'] = "test"
block2 = BlockA()
block2['file'] = "other test"
Now, you've got a guarantee that all instances of your first block object will have the same keys and no additional keys. You can make similar classes for your other blocks, or some general class, or some mix of the two using inheritance. Now to make your data structure:
blocks = [ [block1, block2], [], [], [] ]
print(blocks) # Or "print blocks" if you're not using Python 3.x
blocks[0][0]['file'] = "some new file"
print(blocks)
It might also be worthwhile to have a class for this blocks container, with specific methods for adding blocks of each type and accessing blocks of each type. That way you wouldn't trip yourself up with accidentally adding the wrong kind of block to one of the four lists or similar issues. But depending on how much you'll be using this structure, that could be overkill.

Related

how to creat multiple empty dictionaries from names in a list in python?

I need to create multiple dictionaries from a list. if list is ['dic1','dict2'], i want to creat two different dictionaries such as sample_dic1 and sample_dic2.
if I don't use loops, I'll just type:
sample_dic1=dict();sample_dic2=dict()
my question is how to do it in a loop from a name list.
I tried to put the list in loop while each value of the loop equal to dict().
However, it does not assign the left-hand side to dict().
di_list=['dic1','dict2']
for (a) in di_list:
'sample_{}'.format(a)=dict()
I also Tried this. it doesn't give any error. but doesn't work neigher
temp=dict()
di_list=['dic1','dict2']
for (a) in di_list:
temp[a]='sample_{}'.format(a)
temp[a]=dict()
so I want to creat these two dictrionaries from di_list values. with 'sample_{}'.format(a) I can creat my desired name, But binding it to dict() doesn't work. i.e,. sample_{}'.format(a)=dict()
I think you're mixing two things: variables names and values. Variable names do not have a real effective meaning and if you consistently change their names the program remains the same (in fact, it happens in most languages under the hood anyway).
Here is an option to refer to an arbitrary number of values by name, using a dictionary (name->value):
temp=dict()
di_list=['dic1','dict2']
for (a) in di_list:
temp['sample_{}'.format(a)] = dict()
Now you can verify the values are in fact there:
assert temp['sample_dic1'] == {} # True
assert temp['sample_dict2'] == {} # True

Duplicating without referencing in python

How can i duplicate a list of lists (or any other types) in a way that the resulting lists are new objects and not references to the old ones? As an example i have the following list of lists:
l=[[1,2],[3,4]]
what i want as result is:
l=[[1,2],[3,4],[1,2],[3,4]]
If i do l*=2 the new sub-lists are references to the old sub-lists.
Doing l[0].append("python") will result in
l=[[1,2,'python'],[3,4],[1,2,'python'],[3,4]]
Also creating a new list like:
l2=list(l)
or
l2=l[:]
doesn't solve the problem. I want to have new sub-lists which are independent of their origin and which upon changing have no impact on their old fellows. How can i do this i python?
In general, the best way to copy a nested data structure so that copies get made of all the references (not just the ones at the top level) is to use copy.deepcopy. In your nested list example, you can do:
l.extend(copy.deepcopy(l))
deepcopy will still work even if the data structure contains references to itself, or multiple references to the same object. It usually works for objects stored as attributes on an instances of custom classes too. You can define a __deepcopy__ method if you want to give a class special copying behavior (e.g. if some of its attributes are bookkeeping data that shouldn't be copied).
Here's a version of your nested list example code using instances of a linked list class rather than Python lists. copy.deepcopy does the right thing!
class linked_list(object):
def __init__(self, value, next=None):
self.value = value
self.next = next
def __repr__(self):
if self.next is not None:
return "({!r})->{!r}".format(self.value, self.next)
else:
return "({!r})".format self.value
lst = linked_list(linked_list(1, linked_list(2)),
linked_list(linked_list(3, linked_list(4))))
print(lst) # prints ((1)->(2))->((3)->(4))
lst.next.next = copy.deepcopy(lst)
print(lst) # prints ((1)->(2))->((3)->(4))->((1)->(2))->((3)->(4))
lst.value.value = 5
print(lst) # prints ((5)->(2))->((3)->(4))->((1)->(2))->((3)->(4))

Reference a list of dicts

Python 2.7 on Mint Cinnamon 17.3.
I have a bit of test code employing a list of dicts and despite many hours of frustration, I cannot seem to work out why it is not working as it should do.
blockagedict = {'location': None, 'timestamp': None, 'blocked': None}
blockedlist = [blockagedict]
blockagedict['location'] = 'A'
blockagedict['timestamp'] = '12-Apr-2016 01:01:08.702149'
blockagedict['blocked'] = True
blockagedict['location'] = 'B'
blockagedict['timestamp'] = '12-Apr-2016 01:01:09.312459'
blockagedict['blocked'] = False
blockedlist.append(blockagedict)
for test in blockedlist:
print test['location'], test['timestamp'], test['blocked']
This always produces the following output and I cannot work out why and cannot see if I have anything wrong with my code. It always prints out the last set of dict values but should print all, if I am not mistaken.
B 12-Apr-2016 01:01:09.312459 False
B 12-Apr-2016 01:01:09.312459 False
I would be happy for someone to show me the error of my ways and put me out of my misery.
It is because the line blockedlist = [blockagedict] actually stores a reference to the dict, not a copy, in the list. Your code effectively creates a list that has two references to the very same object.
If you care about performance and will have 1 million dictionaries in a list, all with the same keys, you will be better off using a NumPy structured array. Then you can have a single, efficient data structure which is basically a matrix of rows and named columns of appropriate types. You mentioned in a comment that you may know the number of rows in advance. Here's a rewrite of your example code using NumPy instead, which will be massively more efficient than a list of a million dicts.
import numpy as np
dtype = [('location', str, 1), ('timestamp', str, 27), ('blocked', bool)]
count = 2 # will be much larger in the real program
blockages = np.empty(count, dtype) # use zeros() instead if some data may never be populated
blockages[0]['location'] = 'A'
blockages[0]['timestamp'] = '12-Apr-2016 01:01:08.702149'
blockages[0]['blocked'] = True
blockages['location'][1] = 'B' # n.b. indexing works this way too
blockages['timestamp'][1] = '12-Apr-2016 01:01:09.312459'
blockages['blocked'][1] = False
for test in blockages:
print test['location'], test['timestamp'], test['blocked']
Note that the usage is almost identical. But the storage is in a fixed size, single allocation. This will reduce memory usage and compute time.
As a nice side effect, writing it as above completely sidesteps the issue you originally had, with multiple references to the same row. Now all the data is placed directly into the matrix with no object references at all.
Later in a comment you mention you cannot use NumPy because it may not be installed. Well, we can still avoid unnecessary dicts, like this:
from array import array
blockages = {'location': [], 'timestamp': [], 'blocked': array('B')}
blockages['location'].append('A')
blockages['timestamp'].append('12-Apr-2016 01:01:08.702149')
blockages['blocked'].append(True)
blockages['location'].append('B')
blockages['timestamp'].append('12-Apr-2016 01:01:09.312459')
blockages['blocked'].append(False)
for location, timestamp, blocked in zip(*blockages.values()):
print location, timestamp, blocked
Note I use array here for efficient storage of the fixed-size blocked values (this way each value takes exactly one byte).
You still end up with resizable lists that you could avoid, but at least you don't need to store a dict in every slot of the list. This should still be more efficient.
Ok, I have initialised the list of dicts right off the bat and this seems to work. Although I am tempted to write a class for this.
blockedlist = [{'location': None, 'timestamp': None, 'blocked': None} for k in range(2)]
blockedlist[0]['location'] = 'A'
blockedlist[0]['timestamp'] = '12-Apr-2016 01:01:08.702149'
blockedlist[0]['blocked'] = True
blockedlist[1]['location'] = 'B'
blockedlist[1]['timestamp'] = '12-Apr-2016 01:01:09.312459'
blockedlist[1]['blocked'] = False
for test in blockedlist:
print test['location'], test['timestamp'], test['blocked']
And this produces what I was looking for:
A 12-Apr-2016 01:01:08.702149 True
B 12-Apr-2016 01:01:09.312459 False
I will be reading from a text file with 1 to 2 million lines, so converting the code to iterate through the lines won't be a problem.

Python - null object pattern with generators

It is apparently Pythonic to return values that can be treated as 'False' versions of the successful return type, such that if MyIterableObject: do_things() is a simple way to deal with the output whether or not it is actually there.
With generators, bool(MyGenerator) is always True even if it would have a len of 0 or something equally empty. So while I could write something like the following:
result = list(get_generator(*my_variables))
if result:
do_stuff(result)
It seems like it defeats the benefit of having a generator in the first place.
Perhaps I'm just missing a language feature or something, but what is the pythonic language construct for explicitly indicating that work is not to be done with empty generators?
To be clear, I'd like to be able to give the user some insight as to how much work the script actually did (if any) - contextual snippet as follows:
# Python 2.7
templates = files_from_folder(path_to_folder)
result = list(get_same_sections(templates)) # returns generator
if not result:
msg("No data to sync.")
sys.exit()
for data in result:
for i, tpl in zip(data, templates):
tpl['sections'][i]['uuid'] = data[-1]
msg("{} sections found to sync up.".format(len(result)))
It works, but I think that ultimately it's a waste to change the generator into a list just to see if there's any work to do, so I assume there's a better way, yes?
EDIT: I get the sense that generators just aren't supposed to be used in this way, but I will add an example to show my reasoning.
There's a semi-popular 'helper function' in Python that you see now and again when you need to traverse a structure like a nested dict or what-have-you. Usually called getnode or getn, whenever I see it, it reads something like this:
def get_node(seq, path):
for p in path:
if p in seq:
seq = seq[p]
else:
return ()
return seq
So in this way, you can make it easier to deal with the results of a complicated path to data in a nested structure without always checking for None or try/except when you're not actually dealing with 'something exceptional'.
mydata = get_node(my_container, ('path', 2, 'some', 'data'))
if mydata: # could also be "for x in mydata", etc
do_work(mydata)
else:
something_else()
It's looking less like this kind of syntax would (or could) exist with generators, without writing a class that handles generators in this way as has been suggested.
A generator does not have a length until you've exhausted its iterations.
the only way to get whether it's got anything or not, is to exhaust it
items = list(myGenerator)
if items:
# do something
Unless you wrote a class with attribute nonzero that internally looks at your iterations list
class MyGenerator(object):
def __init__(self, items):
self.items = items
def __iter__(self):
for i in self.items:
yield i
def __nonzero__(self):
return bool(self.items)
>>> bool(MyGenerator([]))
False
>>> bool(MyGenerator([1]))
True
>>>

Iterating over a large unicode list taking a long time?

I'm working with the program Autodesk Maya.
I've made a naming convention script that will name each item in a certain convention accordingly. However I have it list every time in the scene, then check if the chosen name matches any current name in the scene, and then I have it rename it and recheck once more through the scene if there is a duplicate.
However, when i run the code, it can take as long as 30 seconds to a minute or more to run through it all. At first I had no idea what was making my code run slow, as it worked fine in a relatively low scene amount. But then when i put print statements in the check scene code, i saw that it was taking a long time to check through all the items in the scene, and check for duplicates.
The ls() command provides a unicode list of all the items in the scene. These items can be relatively large, up to a thousand or more if the scene has even a moderate amount of items, a normal scene would be several times larger than the testing scene i have at the moment (which has about 794 items in this list).
Is this supposed to take this long? Is the method i'm using to compare things inefficient? I'm not sure what to do here, the code is taking an excessive amount of time, i'm also wondering if it could be anything else in the code, but this seems like it might be it.
Here is some code below.
class Name(object):
"""A naming convention class that runs passed arguments through user
dictionary, and returns formatted string of users input naming convention.
"""
def __init__(self, user_conv):
self.user_conv = user_conv
# an example of a user convention is '${prefix}_${name}_${side}_${objtype}'
#staticmethod
def abbrev_lib(word):
# a dictionary of abbreviated words is here, takes in a string
# and returns an abbreviated string, if not found return given string
#staticmethod
def check_scene(name):
"""Checks entire scene for same name. If duplicate exists,
Keyword Arguments:
name -- (string) name of object to be checked
"""
scene = ls()
match = [x for x in scene if isinstance(x, collections.Iterable)
and (name in x)]
if not match:
return name
else:
return ''
def convert(self, prefix, name, side, objtype):
"""Converts given information about object into user specified convention.
Keyword Arguments:
prefix -- what is prefixed before the name
name -- name of the object or node
side -- what side the object is on, example 'left' or 'right'
obj_type -- the type of the object, example 'joint' or 'multiplyDivide'
"""
prefix = self.abbrev_lib(prefix)
name = self.abbrev_lib(name)
side = ''.join([self.abbrev_lib(x) for x in side])
objtype = self.abbrev_lib(objtype)
i = 02
checked = ''
subs = {'prefix': prefix, 'name': name, 'side':
side, 'objtype': objtype}
while self.checked == '':
newname = Template (self.user_conv.lower())
newname = newname.safe_substitute(**subs)
newname = newname.strip('_')
newname = newname.replace('__', '_')
checked = self.check_scene(newname)
if checked == '' and i < 100:
subs['objtype'] = '%s%s' %(objtype, i)
i+=1
else:
break
return checked
are you running this many times? You are potentially trolling a list of several hundred or a few thousand items for each iteration inside while self.checked =='', which would be a likely culprit. FWIW prints are also very slow in Maya, especially if you're printing a long list - so doing that many times will definitely be slow no matter what.
I'd try a couple of things to speed this up:
limit your searches to one type at a time - why troll through hundreds of random nodes if you only care about MultiplyDivide right now?
Use a set or a dictionary to search, rather than a list - sets and dictionaries use hashsets and are faster for lookups
If you're worried about maintining a naming convetion, definitely design it to be resistant to Maya's default behavior which is to append numeric suffixes to keep names unique. Any naming convention which doesn't support this will be a pain in the butt for all time, because you can't prevent Maya from doing this in the ordinary course of business. On the other hand if you use that for differntiating instances you don't need to do any uniquification at all - just use rename() on the object and capture the result. The weakness there is that Maya won't rename for global uniqueness, only local - so if you want to make unique node name for things that are not siblings you have to do it yourself.
Here's some cheapie code for finding unique node names:
def get_unique_scene_names (*nodeTypes):
if not nodeTypes:
nodeTypes = ('transform',)
results = {}
for longname in cmds.ls(type = nodeTypes, l=True):
shortname = longname.rpartition("|")[-1]
if not shortname in results:
results[shortname] = set()
results[shortname].add(longname)
return results
def add_unique_name(item, node_dict):
shortname = item.rpartition("|")[-1]
if shortname in node_dict:
node_dict[shortname].add(item)
else:
node_dict[shortname] = set([item])
def remove_unique_name(item, node_dict):
shortname = item.rpartition("|")[-1]
existing = node_dict.get(shortname, [])
if item in existing:
existing.remove(item)
def apply_convention(node, new_name, node_dict):
if not new_name in node_dict:
renamed_item = cmds.ls(cmds.rename(node, new_name), l=True)[0]
remove_unique_name(node, node_dict)
add_unique_name ( renamed_item, node_dict)
return renamed_item
else:
for n in range(99999):
possible_name = new_name + str(n + 1)
if not possible_name in node_dict:
renamed_item = cmds.ls(cmds.rename(node, possible_name), l=True)[0]
add_unique_name(renamed_item, node_dict)
return renamed_item
raise RuntimeError, "Too many duplicate names"
To use it on a particular node type, you just supply the right would-be name when calling apply_convention(). This would rename all the joints in the scene (naively!) to 'jnt_X' while keeping the suffixes unique. You'd do something smarter than that, like your original code did - this just makes sure that leaves are unique:
joint_names= get_unique_scene_names('joint')
existing = cmds.ls( type='joint', l = True)
existing .sort()
existing .reverse()
# do this to make sure it works from leaves backwards!
for item in existing :
apply_convention(item, 'jnt_', joint_names)
# check the uniqueness constraint by looking for how many items share a short name in the dict:
for d in joint_names:
print d, len (joint_names[d])
But, like i said, plan for those damn numeric suffixes, maya makes them all the time without asking for permission so you can't fight em :(
Instead of running ls for each and every name, you should run it once and store that result into a set (an unordered list - slightly faster). Then check against that when you run check_scene
def check_scene(self, name):
"""Checks entire scene for same name. If duplicate exists,
Keyword Arguments:
name -- (string) name of object to be checked
"""
if not hasattr(self, 'scene'):
self.scene = set(ls())
if name not in self.scene:
return name
else:
return ''