How do I sort a list of dictionaries by a value of the dictionary?
Question:
How do I sort a list of dictionaries by a specific key’s value? Given:
[{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
when sorted by name, it should become:
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
Answers:
You have to implement your own comparison function that will compare the dictionaries by values of name keys. See Sorting Mini-HOW TO from PythonInfo Wiki
I guess you’ve meant:
[{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
This would be sorted like this:
sorted(l,cmp=lambda x,y: cmp(x['name'],y['name']))
my_list = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
my_list.sort(lambda x,y : cmp(x['name'], y['name']))
my_list will now be what you want.
Or better:
Since Python 2.4, there’s a key argument is both more efficient and neater:
my_list = sorted(my_list, key=lambda k: k['name'])
…the lambda is, IMO, easier to understand than operator.itemgetter, but your mileage may vary.
The sorted() function takes a key= parameter
newlist = sorted(list_to_be_sorted, key=lambda d: d['name'])
Alternatively, you can use operator.itemgetter instead of defining the function yourself
from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name'))
For completeness, add reverse=True to sort in descending order
newlist = sorted(list_to_be_sorted, key=itemgetter('name'), reverse=True)
import operator
a_list_of_dicts.sort(key=operator.itemgetter('name'))
‘key’ is used to sort by an arbitrary value and ‘itemgetter’ sets that value to each item’s ‘name’ attribute.
You could use a custom comparison function, or you could pass in a function that calculates a custom sort key. That’s usually more efficient as the key is only calculated once per item, while the comparison function would be called many more times.
You could do it this way:
def mykey(adict): return adict['name']
x = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age':10}]
sorted(x, key=mykey)
But the standard library contains a generic routine for getting items of arbitrary objects: itemgetter. So try this instead:
from operator import itemgetter
x = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age':10}]
sorted(x, key=itemgetter('name'))
import operator
To sort the list of dictionaries by key=’name’:
list_of_dicts.sort(key=operator.itemgetter('name'))
To sort the list of dictionaries by key=’age’:
list_of_dicts.sort(key=operator.itemgetter('age'))
If you want to sort the list by multiple keys, you can do the following:
my_list = [{'name':'Homer', 'age':39}, {'name':'Milhouse', 'age':10}, {'name':'Bart', 'age':10} ]
sortedlist = sorted(my_list , key=lambda elem: "%02d %s" % (elem['age'], elem['name']))
It is rather hackish, since it relies on converting the values into a single string representation for comparison, but it works as expected for numbers including negative ones (although you will need to format your string appropriately with zero paddings if you are using numbers).
Using the Schwartzian transform from Perl,
py = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
do
sort_on = "name"
decorated = [(dict_[sort_on], dict_) for dict_ in py]
decorated.sort()
result = [dict_ for (key, dict_) in decorated]
gives
>>> result
[{'age': 10, 'name': 'Bart'}, {'age': 39, 'name': 'Homer'}]
More on the Perl Schwartzian transform:
In computer science, the Schwartzian transform is a Perl programming
idiom used to improve the efficiency of sorting a list of items. This
idiom is appropriate for comparison-based sorting when the ordering is
actually based on the ordering of a certain property (the key) of the
elements, where computing that property is an intensive operation that
should be performed a minimal number of times. The Schwartzian
Transform is notable in that it does not use named temporary arrays.
Let’s say I have a dictionary D with the elements below. To sort, just use the key argument in sorted to pass a custom function as below:
D = {'eggs': 3, 'ham': 1, 'spam': 2}
def get_count(tuple):
return tuple[1]
sorted(D.items(), key = get_count, reverse=True)
# Or
sorted(D.items(), key = lambda x: x[1], reverse=True) # Avoiding get_count function call
Check this out.
Here is the alternative general solution – it sorts elements of a dict by keys and values.
The advantage of it – no need to specify keys, and it would still work if some keys are missing in some of dictionaries.
def sort_key_func(item):
""" Helper function used to sort list of dicts
:param item: dict
:return: sorted list of tuples (k, v)
"""
pairs = []
for k, v in item.items():
pairs.append((k, v))
return sorted(pairs)
sorted(A, key=sort_key_func)
Using the Pandas package is another method, though its runtime at large scale is much slower than the more traditional methods proposed by others:
import pandas as pd
listOfDicts = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
df = pd.DataFrame(listOfDicts)
df = df.sort_values('name')
sorted_listOfDicts = df.T.to_dict().values()
Here are some benchmark values for a tiny list and a large (100k+) list of dicts:
setup_large = "listOfDicts = [];
[listOfDicts.extend(({'name':'Homer', 'age':39}, {'name':'Bart', 'age':10})) for _ in range(50000)];
from operator import itemgetter;import pandas as pd;
df = pd.DataFrame(listOfDicts);"
setup_small = "listOfDicts = [];
listOfDicts.extend(({'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}));
from operator import itemgetter;import pandas as pd;
df = pd.DataFrame(listOfDicts);"
method1 = "newlist = sorted(listOfDicts, key=lambda k: k['name'])"
method2 = "newlist = sorted(listOfDicts, key=itemgetter('name')) "
method3 = "df = df.sort_values('name');
sorted_listOfDicts = df.T.to_dict().values()"
import timeit
t = timeit.Timer(method1, setup_small)
print('Small Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_small)
print('Small Method LC2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup_small)
print('Small Method Pandas: ' + str(t.timeit(100)))
t = timeit.Timer(method1, setup_large)
print('Large Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_large)
print('Large Method LC2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup_large)
print('Large Method Pandas: ' + str(t.timeit(1)))
#Small Method LC: 0.000163078308105
#Small Method LC2: 0.000134944915771
#Small Method Pandas: 0.0712950229645
#Large Method LC: 0.0321750640869
#Large Method LC2: 0.0206089019775
#Large Method Pandas: 5.81405615807
a = [{'name':'Homer', 'age':39}, ...]
# This changes the list a
a.sort(key=lambda k : k['name'])
# This returns a new list (a is not modified)
sorted(a, key=lambda k : k['name'])
Sometime we need to use lower() for case-insensitive sorting. For example,
lists = [{'name':'Homer', 'age':39},
{'name':'Bart', 'age':10},
{'name':'abby', 'age':9}]
lists = sorted(lists, key=lambda k: k['name'])
print(lists)
# Bart, Homer, abby
# [{'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}, {'name':'abby', 'age':9}]
lists = sorted(lists, key=lambda k: k['name'].lower())
print(lists)
# abby, Bart, Homer
# [ {'name':'abby', 'age':9}, {'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}]
If you do not need the original list of dictionaries, you could modify it in-place with sort() method using a custom key function.
Key function:
def get_name(d):
""" Return the value of a key in a dictionary. """
return d["name"]
The list to be sorted:
data_one = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
Sorting it in-place:
data_one.sort(key=get_name)
If you need the original list, call the sorted() function passing it the list and the key function, then assign the returned sorted list to a new variable:
data_two = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
new_data = sorted(data_two, key=get_name)
Printing data_one and new_data.
>>> print(data_one)
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
>>> print(new_data)
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
I have been a big fan of a filter with lambda. However, it is not best option if you consider time complexity.
First option
sorted_list = sorted(list_to_sort, key= lambda x: x['name'])
# Returns list of values
Second option
list_to_sort.sort(key=operator.itemgetter('name'))
# Edits the list, and does not return a new list
Fast comparison of execution times
# First option
python3.6 -m timeit -s "list_to_sort = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Faaa', 'age':57}, {'name':'Errr', 'age':20}]" -s "sorted_l=[]" "sorted_l = sorted(list_to_sort, key=lambda e: e['name'])"
1000000 loops, best of 3: 0.736 µsec per loop
# Second option
python3.6 -m timeit -s "list_to_sort = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Faaa', 'age':57}, {'name':'Errr', 'age':20}]" -s "sorted_l=[]" -s "import operator" "list_to_sort.sort(key=operator.itemgetter('name'))"
1000000 loops, best of 3: 0.438 µsec per loop
If performance is a concern, I would use operator.itemgetter instead of lambda as built-in functions perform faster than hand-crafted functions. The itemgetter function seems to perform approximately 20% faster than lambda based on my testing.
From https://wiki.python.org/moin/PythonSpeed:
Likewise, the builtin functions run faster than hand-built equivalents. For example, map(operator.add, v1, v2) is faster than map(lambda x,y: x+y, v1, v2).
Here is a comparison of sorting speed using lambda vs itemgetter.
import random
import operator
# Create a list of 100 dicts with random 8-letter names and random ages from 0 to 100.
l = [{'name': ''.join(random.choices(string.ascii_lowercase, k=8)), 'age': random.randint(0, 100)} for i in range(100)]
# Test the performance with a lambda function sorting on name
%timeit sorted(l, key=lambda x: x['name'])
13 µs ± 388 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# Test the performance with itemgetter sorting on name
%timeit sorted(l, key=operator.itemgetter('name'))
10.7 µs ± 38.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# Check that each technique produces the same sort order
sorted(l, key=lambda x: x['name']) == sorted(l, key=operator.itemgetter('name'))
True
Both techniques sort the list in the same order (verified by execution of the final statement in the code block), but the first one is a little faster.
As indicated by @Claudiu to @monojohnny in comment section of this answer,
given:
list_to_be_sorted = [
{'name':'Homer', 'age':39},
{'name':'Milhouse', 'age':10},
{'name':'Bart', 'age':10}
]
to sort the list of dictionaries by key 'age', 'name'
(like in SQL statement ORDER BY age, name), you can use:
newlist = sorted( list_to_be_sorted, key=lambda k: (k['age'], k['name']) )
or, likewise
import operator
newlist = sorted( list_to_be_sorted, key=operator.itemgetter('age','name') )
print(newlist)
[{‘name’: ‘Bart’, ‘age’: 10},
{‘name’: ‘Milhouse’, ‘age’: 10},
{‘name’: ‘Homer’, ‘age’: 39}]
sorting by multiple columns, while in descending order on some of them:
the cmps array is global to the cmp function, containing field names and inv == -1 for desc 1 for asc
def cmpfun(a, b):
for (name, inv) in cmps:
res = cmp(a[name], b[name])
if res != 0:
return res * inv
return 0
data = [
dict(name='alice', age=10),
dict(name='baruch', age=9),
dict(name='alice', age=11),
]
all_cmps = [
[('name', 1), ('age', -1)],
[('name', 1), ('age', 1)],
[('name', -1), ('age', 1)],]
print 'data:', data
for cmps in all_cmps: print 'sort:', cmps; print sorted(data, cmpfun)
You can use the following:
lst = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
sorted_lst = sorted(lst, key=lambda x: x['age']) # change this to sort by a different field
print(sorted_lst)
It might be better to use dict.get() to fetch the values to sort by in the sorting key. One way it’s better than dict[] is that a default value may be used to if a key is missing in some dictionary in the list.
For example, if a list of dicts were sorted by 'age' but 'age' was missing in some dict, that dict can either be pushed to the back of the sorted list (or to the front) by simply passing inf as a default value to dict.get().
lst = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}, {'name': 'Lisa'}]
sorted(lst, key=lambda d: d['age']) # KeyError: 'age'
sorted(lst, key=itemgetter('age')) # KeyError: 'age'
# push dicts with missing keys to the back
sorted(lst, key=lambda d: d.get('age', float('inf'))) # OK
# push dicts with missing keys to the front
sorted(lst, key=lambda d: d.get('age', -float('inf'))) # OK
# if the value to be sorted by is a string
# '~' because it has the highest printable ASCII value
sorted(lst, key=lambda d: d.get('name', '~')) # OK
How do I sort a list of dictionaries by a specific key’s value? Given:
[{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
when sorted by name, it should become:
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
You have to implement your own comparison function that will compare the dictionaries by values of name keys. See Sorting Mini-HOW TO from PythonInfo Wiki
I guess you’ve meant:
[{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
This would be sorted like this:
sorted(l,cmp=lambda x,y: cmp(x['name'],y['name']))
my_list = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
my_list.sort(lambda x,y : cmp(x['name'], y['name']))
my_list will now be what you want.
Or better:
Since Python 2.4, there’s a key argument is both more efficient and neater:
my_list = sorted(my_list, key=lambda k: k['name'])
…the lambda is, IMO, easier to understand than operator.itemgetter, but your mileage may vary.
The sorted() function takes a key= parameter
newlist = sorted(list_to_be_sorted, key=lambda d: d['name'])
Alternatively, you can use operator.itemgetter instead of defining the function yourself
from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name'))
For completeness, add reverse=True to sort in descending order
newlist = sorted(list_to_be_sorted, key=itemgetter('name'), reverse=True)
import operator
a_list_of_dicts.sort(key=operator.itemgetter('name'))
‘key’ is used to sort by an arbitrary value and ‘itemgetter’ sets that value to each item’s ‘name’ attribute.
You could use a custom comparison function, or you could pass in a function that calculates a custom sort key. That’s usually more efficient as the key is only calculated once per item, while the comparison function would be called many more times.
You could do it this way:
def mykey(adict): return adict['name']
x = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age':10}]
sorted(x, key=mykey)
But the standard library contains a generic routine for getting items of arbitrary objects: itemgetter. So try this instead:
from operator import itemgetter
x = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age':10}]
sorted(x, key=itemgetter('name'))
import operator
To sort the list of dictionaries by key=’name’:
list_of_dicts.sort(key=operator.itemgetter('name'))
To sort the list of dictionaries by key=’age’:
list_of_dicts.sort(key=operator.itemgetter('age'))
If you want to sort the list by multiple keys, you can do the following:
my_list = [{'name':'Homer', 'age':39}, {'name':'Milhouse', 'age':10}, {'name':'Bart', 'age':10} ]
sortedlist = sorted(my_list , key=lambda elem: "%02d %s" % (elem['age'], elem['name']))
It is rather hackish, since it relies on converting the values into a single string representation for comparison, but it works as expected for numbers including negative ones (although you will need to format your string appropriately with zero paddings if you are using numbers).
Using the Schwartzian transform from Perl,
py = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
do
sort_on = "name"
decorated = [(dict_[sort_on], dict_) for dict_ in py]
decorated.sort()
result = [dict_ for (key, dict_) in decorated]
gives
>>> result
[{'age': 10, 'name': 'Bart'}, {'age': 39, 'name': 'Homer'}]
More on the Perl Schwartzian transform:
In computer science, the Schwartzian transform is a Perl programming
idiom used to improve the efficiency of sorting a list of items. This
idiom is appropriate for comparison-based sorting when the ordering is
actually based on the ordering of a certain property (the key) of the
elements, where computing that property is an intensive operation that
should be performed a minimal number of times. The Schwartzian
Transform is notable in that it does not use named temporary arrays.
Let’s say I have a dictionary D with the elements below. To sort, just use the key argument in sorted to pass a custom function as below:
D = {'eggs': 3, 'ham': 1, 'spam': 2}
def get_count(tuple):
return tuple[1]
sorted(D.items(), key = get_count, reverse=True)
# Or
sorted(D.items(), key = lambda x: x[1], reverse=True) # Avoiding get_count function call
Check this out.
Here is the alternative general solution – it sorts elements of a dict by keys and values.
The advantage of it – no need to specify keys, and it would still work if some keys are missing in some of dictionaries.
def sort_key_func(item):
""" Helper function used to sort list of dicts
:param item: dict
:return: sorted list of tuples (k, v)
"""
pairs = []
for k, v in item.items():
pairs.append((k, v))
return sorted(pairs)
sorted(A, key=sort_key_func)
Using the Pandas package is another method, though its runtime at large scale is much slower than the more traditional methods proposed by others:
import pandas as pd
listOfDicts = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
df = pd.DataFrame(listOfDicts)
df = df.sort_values('name')
sorted_listOfDicts = df.T.to_dict().values()
Here are some benchmark values for a tiny list and a large (100k+) list of dicts:
setup_large = "listOfDicts = [];
[listOfDicts.extend(({'name':'Homer', 'age':39}, {'name':'Bart', 'age':10})) for _ in range(50000)];
from operator import itemgetter;import pandas as pd;
df = pd.DataFrame(listOfDicts);"
setup_small = "listOfDicts = [];
listOfDicts.extend(({'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}));
from operator import itemgetter;import pandas as pd;
df = pd.DataFrame(listOfDicts);"
method1 = "newlist = sorted(listOfDicts, key=lambda k: k['name'])"
method2 = "newlist = sorted(listOfDicts, key=itemgetter('name')) "
method3 = "df = df.sort_values('name');
sorted_listOfDicts = df.T.to_dict().values()"
import timeit
t = timeit.Timer(method1, setup_small)
print('Small Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_small)
print('Small Method LC2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup_small)
print('Small Method Pandas: ' + str(t.timeit(100)))
t = timeit.Timer(method1, setup_large)
print('Large Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_large)
print('Large Method LC2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup_large)
print('Large Method Pandas: ' + str(t.timeit(1)))
#Small Method LC: 0.000163078308105
#Small Method LC2: 0.000134944915771
#Small Method Pandas: 0.0712950229645
#Large Method LC: 0.0321750640869
#Large Method LC2: 0.0206089019775
#Large Method Pandas: 5.81405615807
a = [{'name':'Homer', 'age':39}, ...]
# This changes the list a
a.sort(key=lambda k : k['name'])
# This returns a new list (a is not modified)
sorted(a, key=lambda k : k['name'])
Sometime we need to use lower() for case-insensitive sorting. For example,
lists = [{'name':'Homer', 'age':39},
{'name':'Bart', 'age':10},
{'name':'abby', 'age':9}]
lists = sorted(lists, key=lambda k: k['name'])
print(lists)
# Bart, Homer, abby
# [{'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}, {'name':'abby', 'age':9}]
lists = sorted(lists, key=lambda k: k['name'].lower())
print(lists)
# abby, Bart, Homer
# [ {'name':'abby', 'age':9}, {'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}]
If you do not need the original list of dictionaries, you could modify it in-place with sort() method using a custom key function.
Key function:
def get_name(d):
""" Return the value of a key in a dictionary. """
return d["name"]
The list to be sorted:
data_one = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
Sorting it in-place:
data_one.sort(key=get_name)
If you need the original list, call the sorted() function passing it the list and the key function, then assign the returned sorted list to a new variable:
data_two = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
new_data = sorted(data_two, key=get_name)
Printing data_one and new_data.
>>> print(data_one)
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
>>> print(new_data)
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
I have been a big fan of a filter with lambda. However, it is not best option if you consider time complexity.
First option
sorted_list = sorted(list_to_sort, key= lambda x: x['name'])
# Returns list of values
Second option
list_to_sort.sort(key=operator.itemgetter('name'))
# Edits the list, and does not return a new list
Fast comparison of execution times
# First option
python3.6 -m timeit -s "list_to_sort = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Faaa', 'age':57}, {'name':'Errr', 'age':20}]" -s "sorted_l=[]" "sorted_l = sorted(list_to_sort, key=lambda e: e['name'])"
1000000 loops, best of 3: 0.736 µsec per loop
# Second option
python3.6 -m timeit -s "list_to_sort = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Faaa', 'age':57}, {'name':'Errr', 'age':20}]" -s "sorted_l=[]" -s "import operator" "list_to_sort.sort(key=operator.itemgetter('name'))"
1000000 loops, best of 3: 0.438 µsec per loop
If performance is a concern, I would use operator.itemgetter instead of lambda as built-in functions perform faster than hand-crafted functions. The itemgetter function seems to perform approximately 20% faster than lambda based on my testing.
From https://wiki.python.org/moin/PythonSpeed:
Likewise, the builtin functions run faster than hand-built equivalents. For example, map(operator.add, v1, v2) is faster than map(lambda x,y: x+y, v1, v2).
Here is a comparison of sorting speed using lambda vs itemgetter.
import random
import operator
# Create a list of 100 dicts with random 8-letter names and random ages from 0 to 100.
l = [{'name': ''.join(random.choices(string.ascii_lowercase, k=8)), 'age': random.randint(0, 100)} for i in range(100)]
# Test the performance with a lambda function sorting on name
%timeit sorted(l, key=lambda x: x['name'])
13 µs ± 388 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# Test the performance with itemgetter sorting on name
%timeit sorted(l, key=operator.itemgetter('name'))
10.7 µs ± 38.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# Check that each technique produces the same sort order
sorted(l, key=lambda x: x['name']) == sorted(l, key=operator.itemgetter('name'))
True
Both techniques sort the list in the same order (verified by execution of the final statement in the code block), but the first one is a little faster.
As indicated by @Claudiu to @monojohnny in comment section of this answer,
given:
list_to_be_sorted = [
{'name':'Homer', 'age':39},
{'name':'Milhouse', 'age':10},
{'name':'Bart', 'age':10}
]
to sort the list of dictionaries by key 'age', 'name'
(like in SQL statement ORDER BY age, name), you can use:
newlist = sorted( list_to_be_sorted, key=lambda k: (k['age'], k['name']) )
or, likewise
import operator
newlist = sorted( list_to_be_sorted, key=operator.itemgetter('age','name') )
print(newlist)
[{‘name’: ‘Bart’, ‘age’: 10},
{‘name’: ‘Milhouse’, ‘age’: 10},
{‘name’: ‘Homer’, ‘age’: 39}]
sorting by multiple columns, while in descending order on some of them:
the cmps array is global to the cmp function, containing field names and inv == -1 for desc 1 for asc
def cmpfun(a, b):
for (name, inv) in cmps:
res = cmp(a[name], b[name])
if res != 0:
return res * inv
return 0
data = [
dict(name='alice', age=10),
dict(name='baruch', age=9),
dict(name='alice', age=11),
]
all_cmps = [
[('name', 1), ('age', -1)],
[('name', 1), ('age', 1)],
[('name', -1), ('age', 1)],]
print 'data:', data
for cmps in all_cmps: print 'sort:', cmps; print sorted(data, cmpfun)
You can use the following:
lst = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
sorted_lst = sorted(lst, key=lambda x: x['age']) # change this to sort by a different field
print(sorted_lst)
It might be better to use dict.get() to fetch the values to sort by in the sorting key. One way it’s better than dict[] is that a default value may be used to if a key is missing in some dictionary in the list.
For example, if a list of dicts were sorted by 'age' but 'age' was missing in some dict, that dict can either be pushed to the back of the sorted list (or to the front) by simply passing inf as a default value to dict.get().
lst = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}, {'name': 'Lisa'}]
sorted(lst, key=lambda d: d['age']) # KeyError: 'age'
sorted(lst, key=itemgetter('age')) # KeyError: 'age'
# push dicts with missing keys to the back
sorted(lst, key=lambda d: d.get('age', float('inf'))) # OK
# push dicts with missing keys to the front
sorted(lst, key=lambda d: d.get('age', -float('inf'))) # OK
# if the value to be sorted by is a string
# '~' because it has the highest printable ASCII value
sorted(lst, key=lambda d: d.get('name', '~')) # OK