rel2tree

Convert your list of data into JSON serializable structure.

Motivation

Let's suppose you have a set of data given as a list of dicts:

import json

[
  {"name": "Jane", "city": "New York", "sales": 23},
  {"name": "Joe", "city": "New York", "sales": 11},
  {"name": "Jane", "city": "Chicago", "sales": 21},
  {"name": "Jane", "city": "New York", "sales": 4},
  {"name": "Joe", "city": "New York", "sales": 13},
  {"name": "Joe", "city": "Chicago", "sales": 31},
  {"name": "Jane", "city": "New York", "sales": 7},
]

You may want a nice summary, something like this:

[
  {
    "name": "Jane",
    "cities": [
      {
        "city": "New York",
        "sales": 34
      },
      {
        "city": "Chicago",
        "sales": 21
      }
    ],
    "sum": 55
  },
  {
    "name": "Joe",
    "cities": [
      {
        "city": "New York",
        "sales": 24
      },
      {
        "city": "Chicago",
        "sales": 31
      }
    ],
    "sum": 55
  }
]

This can be done relatively easily by iterating over the data set and building the final structure.

summary = {}
for record in data:
    this_person = summary.setdefault(record["name"], {
        "name": record["name"],
        "cities": {},
        "sum": 0,
    })
    this_person_cities = this_person["cities"].setdefault(record["city"], {
        "city": record["city"],
        "sum": 0,
    })
    this_person_cities["sum"] += record["sales"]
    this_person["sum"] += record["sum"]
summary = list(summary.values())
for person in summary:
    person["cities"] = list(person["cities"].values())

print(json.dumps(summary))

Although the above code works well, but it has some problems.

• Not declarative: by looking at the code it is not trivial to tell the final data structure.
• Error-prone.
• The complexity grows with more complex business logic or by adding an additional level.
• Not reusable.

Let's see how you do it with rel2tree:

from rel2tree import f  # NOQA

summary = f.groupby(lambda x: x["name"], f.dict({
    "name": f.groupkey(),
    "cities": f.groupby(lambda x: x["city"], f.dict({
        "city": f.groupkey(),
        "sum": f.map(lambda x: x["sales"]).t(sum)
    })),
    "sum": f.map(lambda x: x["sales"]).t(sum)
}))

print(json.dumps(summary(data)))

Tutorial

map, sort, filter, distinct

The only object one can import from rel2tree is f, which is of type F so we will call it an F object. f is callable, but - on it's own does nothing:

print(f(2))
# 2

Let's say we have a list of numbers (numbers) and we want to duplicate all of it's elements. This can be done in many ways:

• using a list comprehension:

  out = [2 * x for x in numbers]

• using map:

  out = map(lambda x: 2 * x, numbers)

• defining a function (for reusability)

  import functools
  dup = functools.partial(map, lambda x: 2 * x)
  out = dup(numbers)

Using an f it looks like this:

numbers = range(15)
dup = f.map(lambda x: 2 * x)
out = dup(numbers)

This simply made our third approach a little more terse.

Now what if our task is to add 1 to each element after duplication? Can we reuse our dup function? As the result of f.map has the same type as f, we can use map again:

dupplus1 = dup.map(lambda x: x + 1)

f.sort(fnc) sorts our list based on the value of fnc applied to the items (just as the key argument of python's) sorted. f.filter(fnc) keeps only those i items, where fnc(i) is ture(ish). These methods also return Fs (internally the type of f is F) so they are chainable. The F below first duplicates, then filters out big numbers and finally sorts them. (f.sort, without a function sorts the elements.)

f.map(lambda x: 2 * x).filter(lambda x: x < 10).sort()

dict

Back to our numbers, but with the desired output of

{
  "even": [0, 2, 4, 6, 8, 10, 12, 14],
  "odd": [1, 3, 5, 7, 9, 11, 13]
}

We can combine the dict method to achive this:

summary = f.dict({
    "even": f.filter(lambda x: (x % 2 == 0)),
    "odd": f.filter(lambda x: (x % 2 == 1)),
})

If the dictionary values are F objects, those objects will be called with the input list to form the final values, otherwise the values will be left as is.

groupby

To generalize the above example, we can group our numbers based on the remainder devided by, say, 3:

summary = f.groupby(lambda x: x % 3)
# [[0, 3, 6, 9, 12], [1, 4, 7, 10, 13], [2, 5, 8, 11, 14]]

To make it more informative, the desired output should be:

[
  { "remainder": 0, "numbers": [0, 3, 6, 9, 12] },
  { "remainder": 1, "numbers": [1, 4, 7, 10, 13] },
  { "remainder": 2, "numbers": [2, 5, 8, 11, 14] }
]

This can be done by using groupkey:

summary = f.groupby(lambda x: x % 3, f.dict({
  "remainder": f.groupkey(),
  "numbers": f
}))

f.groupkey(level=0) gives the deepest level group key, while f.groupkey(1) is the one level above group key in case of nested groupby's.