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Ch3steR
Ch3steR
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You can avoid assemble, disassemble, tag_datum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom function for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'

You can avoid assemble, disassemble, tag_datum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'

You can avoid assemble, disassemble, tag_datum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom function for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'
deleted 1 character in body
Source Link
Ch3steR
Ch3steR
  • 644
  • 7
  • 14

You can avoid assemble, disassemble, tag_dataumtag_datum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'

You can avoid assemble, disassemble, tag_dataum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'

You can avoid assemble, disassemble, tag_datum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'
added 74 characters in body
Source Link
Ch3steR
Ch3steR
  • 644
  • 7
  • 14

You can avoid cardinality_groups, assemble, disassemble, tag_dataum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'

You can avoid cardinality_groups, assemble, disassemble, tag_dataum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]},
#  {'ids': [4]},
#  {'ids': [3]},
#  {'ids': [2]},
#  {'ids': [3, 4]},
#  {'ids': [1, 2]}]

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'

You can avoid assemble, disassemble, tag_dataum if you leverage on key param of both itertools.groupby and sorted and condense them into one simple function.

Build a custom for key param

Write a function to return the length of each value in the dictionary.

val_len = lambda x: len(x['ids'])

Now pass this as argument for key in both itertools.groupby and sorted

from itertools import groupby

def transform(data):
    sorted_data = sorted(data, key=val_len)
    return [list(group) for _, group in groupby(sorted_data, key=val_len)]

transform(data) # data taken from question itself.
# [[{'ids': [1]}, {'ids': [4]}, {'ids': [3]}, {'ids': [2]}],
#  [{'ids': [3, 4]}, {'ids': [1, 2]}]]

Details

sorted_data = sorted(data, key=val_len) # sorts data based the length of
                                        # the value of 'ids'
# [{'ids': [1]}, ---|
#  {'ids': [4]},    |-- group with length 1
#  {'ids': [3]},    |
#  {'ids': [2]}, ---|

#  {'ids': [3, 4]},---|-- group with length 2
#  {'ids': [1, 2]}]---|

groupby(sorted_data, key=val_len) # groups data based on the length of 
                                  # the value of 'ids'
Source Link
Ch3steR
Ch3steR
  • 644
  • 7
  • 14