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1

1

I need the ability to breadth-first cycle through the deep-nested properties of a dict. This gets me part of the way there (https://stackoverflow.com/a/10756615/5932433) but the result is not what I need.

Given the following data structure:

data = 
 "a": 
 "c": 1,
 "d": 3,
 ,
 "b": 
 "e": 2,
 "f": 4,
 

I need a method that will return the following:

for _, v in cycle(tree_iter(data)):
 print v

# 1 (a -> c)
# 2 (b -> e)
# 3 (a -> d)
# 4 (b -> f)
# 1
# 2
# 3
# 4
# ...etc...

This is the method I have right now for tree_iter:

def tree_iter(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 for inner_key, inner_value in tree_iter(value):
 yield inner_key, inner_value
 else:
 yield key, value

Note that order doesn't need to be guaranteed, as long as it's consistent. Each iteration should cycle through a/b, then cycle through the nested values.

python algorithm iterator

share|improve this question

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also, the structure of the dict will be consistent. All the values will be at level n.
  
  – thejohnbackes
  Mar 26 at 0:45
  

  
  
  
  

add a comment | 

1

1

I need the ability to breadth-first cycle through the deep-nested properties of a dict. This gets me part of the way there (https://stackoverflow.com/a/10756615/5932433) but the result is not what I need.

Given the following data structure:

data = 
 "a": 
 "c": 1,
 "d": 3,
 ,
 "b": 
 "e": 2,
 "f": 4,
 

I need a method that will return the following:

for _, v in cycle(tree_iter(data)):
 print v

# 1 (a -> c)
# 2 (b -> e)
# 3 (a -> d)
# 4 (b -> f)
# 1
# 2
# 3
# 4
# ...etc...

This is the method I have right now for tree_iter:

def tree_iter(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 for inner_key, inner_value in tree_iter(value):
 yield inner_key, inner_value
 else:
 yield key, value

Note that order doesn't need to be guaranteed, as long as it's consistent. Each iteration should cycle through a/b, then cycle through the nested values.

python algorithm iterator

share|improve this question

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also, the structure of the dict will be consistent. All the values will be at level n.
  
  – thejohnbackes
  Mar 26 at 0:45
  

  
  
  
  

add a comment | 

I need the ability to breadth-first cycle through the deep-nested properties of a dict. This gets me part of the way there (https://stackoverflow.com/a/10756615/5932433) but the result is not what I need.

Given the following data structure:

data = 
 "a": 
 "c": 1,
 "d": 3,
 ,
 "b": 
 "e": 2,
 "f": 4,
 

I need a method that will return the following:

for _, v in cycle(tree_iter(data)):
 print v

# 1 (a -> c)
# 2 (b -> e)
# 3 (a -> d)
# 4 (b -> f)
# 1
# 2
# 3
# 4
# ...etc...

This is the method I have right now for tree_iter:

def tree_iter(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 for inner_key, inner_value in tree_iter(value):
 yield inner_key, inner_value
 else:
 yield key, value

Note that order doesn't need to be guaranteed, as long as it's consistent. Each iteration should cycle through a/b, then cycle through the nested values.

python algorithm iterator

share|improve this question

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

data = 
 "a": 
 "c": 1,
 "d": 3,
 ,
 "b": 
 "e": 2,
 "f": 4,
 

for _, v in cycle(tree_iter(data)):
 print v

# 1 (a -> c)
# 2 (b -> e)
# 3 (a -> d)
# 4 (b -> f)
# 1
# 2
# 3
# 4
# ...etc...

def tree_iter(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 for inner_key, inner_value in tree_iter(value):
 yield inner_key, inner_value
 else:
 yield key, value

share|improve this question

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

share|improve this question

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

asked Mar 26 at 0:21

thejohnbackesthejohnbackes

92866 silver badges1616 bronze badges

1 Answer
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0

Your current code appears to do a DFS, so you can return a list of lists from the DFS, zip them, and then flatten. Not the most elegant, but it should work.

def tree_iter_dfs(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 yield value.items()
 else:
 yield [(key, value)]

# https://stackoverflow.com/a/952952/5309823
def flatten(l):
 return [item for sublist in l for item in sublist]

def tree_iter_bfs(nested):
 dfs = tree_iter_dfs(nested)
 return flatten(zip(*dfs))

print(list(tree_iter_bfs(data)))
# [('c', 1), ('e', 2), ('d', 3), ('f', 4)]

Change things around to iterables as necessary; I didn't know what version of Python you were using, etc.

share|improve this answer

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges

add a comment | 

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0

Your current code appears to do a DFS, so you can return a list of lists from the DFS, zip them, and then flatten. Not the most elegant, but it should work.

def tree_iter_dfs(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 yield value.items()
 else:
 yield [(key, value)]

# https://stackoverflow.com/a/952952/5309823
def flatten(l):
 return [item for sublist in l for item in sublist]

def tree_iter_bfs(nested):
 dfs = tree_iter_dfs(nested)
 return flatten(zip(*dfs))

print(list(tree_iter_bfs(data)))
# [('c', 1), ('e', 2), ('d', 3), ('f', 4)]

Change things around to iterables as necessary; I didn't know what version of Python you were using, etc.

share|improve this answer

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges

add a comment | 

0

Your current code appears to do a DFS, so you can return a list of lists from the DFS, zip them, and then flatten. Not the most elegant, but it should work.

def tree_iter_dfs(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 yield value.items()
 else:
 yield [(key, value)]

# https://stackoverflow.com/a/952952/5309823
def flatten(l):
 return [item for sublist in l for item in sublist]

def tree_iter_bfs(nested):
 dfs = tree_iter_dfs(nested)
 return flatten(zip(*dfs))

print(list(tree_iter_bfs(data)))
# [('c', 1), ('e', 2), ('d', 3), ('f', 4)]

Change things around to iterables as necessary; I didn't know what version of Python you were using, etc.

share|improve this answer

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges

add a comment | 

Your current code appears to do a DFS, so you can return a list of lists from the DFS, zip them, and then flatten. Not the most elegant, but it should work.

def tree_iter_dfs(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 yield value.items()
 else:
 yield [(key, value)]

# https://stackoverflow.com/a/952952/5309823
def flatten(l):
 return [item for sublist in l for item in sublist]

def tree_iter_bfs(nested):
 dfs = tree_iter_dfs(nested)
 return flatten(zip(*dfs))

print(list(tree_iter_bfs(data)))
# [('c', 1), ('e', 2), ('d', 3), ('f', 4)]

Change things around to iterables as necessary; I didn't know what version of Python you were using, etc.

share|improve this answer

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges

def tree_iter_dfs(nested):
 for key, value in nested.iteritems():
 if isinstance(value, Mapping):
 yield value.items()
 else:
 yield [(key, value)]

# https://stackoverflow.com/a/952952/5309823
def flatten(l):
 return [item for sublist in l for item in sublist]

def tree_iter_bfs(nested):
 dfs = tree_iter_dfs(nested)
 return flatten(zip(*dfs))

print(list(tree_iter_bfs(data)))
# [('c', 1), ('e', 2), ('d', 3), ('f', 4)]

share|improve this answer

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges

answered Mar 26 at 0:46

colecole

16511 silver badge1010 bronze badges