Leopards is a way to query list of dictionaries or objects as if you are filtering in DBMS. You can get dicts/objects that are matched by OR, AND or NOT or all of them. As you can see in the comparison they are much faster than Pandas.

pip install leopards

from leopards import Q
l = [{"name":"John","age":"16"}, {"name":"Mike","age":"19"},{"name":"Sarah","age":"21"}]
filtered= Q(l,{'name__contains':"k", "age__lt":20})
print(list(filtered))

output

[{'name': 'Mike', 'age': '19'}]

The above filtration can be written as

from leopards import Q

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"}]
filtered = Q(l, name__contains="k", age__lt=20)

Notes:

1. Q returns an iterator which can be converted to a list by calling list.
2. Even though, age was str in the dict, as the value of in the query dict was int, Leopards converted the value in dict automatically to match the query data type. This behaviour can be stopped by passing False to convert_types parameter.

• eq: equals and this default filter
• gt: greater than.
• gte: greater than or equal.
• lt: less than
• lte: less than or equal
• in: the value in a list of a tuple.
  • e.g. age__in=[10,20,30]
• contains: contains a substring as in the example.
• icontains: case-insensitive contains.
• startswith: checks if a value starts with a query strings.
• istartswith: case-insensitive startswith.
• endswith: checks if a value ends with a query strings.
• iendswith: case-insensitive endswith.
• isnull: checks if the value matches any of NULL_VALUES which are ('', '.', None, "None", "null", "NULL")
  • e.g. filter__isnull=True or filter__isnull=False

For eq,gt,gte,lt,lte, in, contains, icontains, startswith,istartswith, endswith and iendswith, you can add a n to negate the results. e.g nin which is equivalent to not in

This section will cover the use of OR, AND and NOT

OR or __or__ takes a list of dictionaries to evaluate and returns with the first True.

from leopards import Q

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"}]
filtered = Q(l, {"OR": [{"name__contains": "k"}, {"age__gte": 21}]})
print(list(filtered))

output

[{'name': 'Mike', 'age': '19'}, {'name': 'Sarah', 'age': '21'}]

NOT or __not__ takes a dict for query run.

from leopards import Q

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"}]
filtered = Q(l, {"age__gt": 15, "NOT": {"age__eq": 19}})
print(list(filtered))

output

[{'name': 'John', 'age': '16'}, {'name': 'Sarah', 'age': '21'}]

AND or __and__ takes a list of dict for query run, returns with the first False.

from leopards import Q

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"}]
filtered = Q(l, {"__and__": [{"age__gte": 15}, {"age__lt": 21}]})
print(list(filtered))

output

[{'name': 'John', 'age': '16'}, {'name': 'Mike', 'age': '19'}]

You can run the following aggregations

• Count
• Max
• Min
• Sum
• Avg

Find the count of certain aggregated column

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"},{"name":"John","age":"19"}]
from leopards import Count
count = Count(l,['age'])

output

[{"age":"16","count":1},{"age":"19","count":2}, {"age":"21","count":1}]

Find the Max value for a certain column in certain aggregated columns

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"},{"name":"John","age":"19"}]
from leopards import Max
count = Max(l,"age",['name'],dtype=int)

output

[{'name': 'John', 'age': '19'}, {'name': 'Mike', 'age': '19'}, {'name': 'Sarah', 'age': '21'}]

Notes:

• If you don't pass the aggregation columns, the maximum will be found across dataset.
• You can pass the datatype of the column to convert it on the fly while evaluating

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"},{"name":"John","age":"19"}]
from leopards import Max
m = Max(l,"age",dtype=int)

output

[{'age': 21}]

Find the Max value for a certain column in certain aggregated columns

l = [{"name": "John", "age": "16"}, {"name": "Mike", "age": "19"}, {"name": "Sarah", "age": "21"},{"name":"John","age":"19"}]
from leopards import Min
m = Min(l,"age",['name'])

output

[{'name': 'John', 'age': '16'}, {'name': 'Mike', 'age': '19'}, {'name': 'Sarah', 'age': '21'}]

Note:

• If you don't pass the aggregation columns, the min will be found across dataset.
• You can pass the datatype of the column to convert it on the fly while evaluating

Like Min and Max but only works with integers and floats.

This is done on Python 3.8 running on Ubuntu 22.04 on i7 11th generation and 32 GB of RAM.

^[1] This was loading the whole csv in memory which was for sake of fair comparison. Nevertheless, Leopards can work with DictReader as an iterable which executes in 0.014s, then it handles line by line.

Thanks for Asma Tahir for Pandas stats.

• saeedesmaili

• Work on CSV Files with Leopards