How do operator.itemgetter() and sort() work?
Question:
I have the following code:
# initialize
a = []
# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 22, "Car Dealer"])
a.append(["Mark", 66, "Retired"])
# sort the table by age
import operator
a.sort(key=operator.itemgetter(1))
# print the table
print(a)
It creates a 4×3 table and then it sorts it by age. My question is, what exactly key=operator.itemgetter(1) does? Does the operator.itemgetter function return the item’s value? Why can’t I just type something like key=a[x][1] there? Or can I? How could with operator print a certain value of the form like 3x2 which is 22?
-
How does exactly Python sort the table? Can I reverse-sort it?
-
How can I sort it based on two columns like first age, and then if age is the same b name?
-
How could I do it without operator?
Answers:
You are asking a lot of questions that you could answer yourself by reading the documentation, so I’ll give you a general advice: read it and experiment in the python shell. You’ll see that itemgetter returns a callable:
>>> func = operator.itemgetter(1)
>>> func(a)
['Paul', 22, 'Car Dealer']
>>> func(a[0])
8
To do it in a different way, you can use lambda:
a.sort(key=lambda x: x[1])
And reverse it:
a.sort(key=operator.itemgetter(1), reverse=True)
Sort by more than one column:
a.sort(key=operator.itemgetter(1,2))
See the sorting How To.
Looks like you’re a little bit confused about all that stuff.
operator is a built-in module providing a set of convenient operators. In two words operator.itemgetter(n) constructs a callable that assumes an iterable object (e.g. list, tuple, set) as input, and fetches the n-th element out of it.
So, you can’t use key=a[x][1] there, because python has no idea what x is. Instead, you could use a lambda function (elem is just a variable name, no magic there):
a.sort(key=lambda elem: elem[1])
Or just an ordinary function:
def get_second_elem(iterable):
return iterable[1]
a.sort(key=get_second_elem)
So, here’s an important note: in python functions are first-class citizens, so you can pass them to other functions as a parameter.
Other questions:
- Yes, you can reverse sort, just add
reverse=True: a.sort(key=..., reverse=True)
- To sort by more than one column you can use
itemgetter with multiple indices: operator.itemgetter(1,2), or with lambda: lambda elem: (elem[1], elem[2]). This way, iterables are constructed on the fly for each item in list, which are than compared against each other in lexicographic(?) order (first elements compared, if equal – second elements compared, etc)
- You can fetch value at [3,2] using
a[2,1] (indices are zero-based). Using operator… It’s possible, but not as clean as just indexing.
Refer to the documentation for details:
Answer for Python beginners
In simpler words:
- The
key= parameter of sort requires a key function (to be applied to be objects to be sorted) rather than a single key value and
- that is just what
operator.itemgetter(1) will give you: A function that grabs the first item from a list-like object.
(More precisely those are callables, not functions, but that is a difference that can often be ignored.)
#sorting first by age then profession,you can change it in function "fun".
a = []
def fun(v):
return (v[1],v[2])
# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])
a.sort(key=fun)
print a
a = []
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])
print a
[['Nick', 30, 'Doctor'], ['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Mark', 66, 'Retired']]
def _cmp(a,b):
if a[1]<b[1]:
return -1
elif a[1]>b[1]:
return 1
else:
return 0
sorted(a,cmp=_cmp)
[['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Nick', 30, 'Doctor'], ['Mark', 66, 'Retired']]
def _key(list_ele):
return list_ele[1]
sorted(a,key=_key)
[['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Nick', 30, 'Doctor'], ['Mark', 66, 'Retired']]
>>>
The easiest way to sort an array using a user-defined function is to use cmp_to_key from Functools.
here is a sample code:
from functools import cmp_to_key
def mine(x,y):
if(x[1]!=y[1]): return x[1]>y[1]
else: return x[0]>y[0]
a = []
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 22, "Car Dealer"])
a.append(["Mark", 66, "Retired"])
def mine(a,b):
if a[1] > b[1]:
return 1
elif a[1] < b[1]:
return -1
else:
if a[0] > b[0]:
return 1
else:
return 0
print(sorted(a,key = cmp_to_key(mine)))
I have the following code:
# initialize
a = []
# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 22, "Car Dealer"])
a.append(["Mark", 66, "Retired"])
# sort the table by age
import operator
a.sort(key=operator.itemgetter(1))
# print the table
print(a)
It creates a 4×3 table and then it sorts it by age. My question is, what exactly key=operator.itemgetter(1) does? Does the operator.itemgetter function return the item’s value? Why can’t I just type something like key=a[x][1] there? Or can I? How could with operator print a certain value of the form like 3x2 which is 22?
-
How does exactly Python sort the table? Can I reverse-sort it?
-
How can I sort it based on two columns like first age, and then if age is the same b name?
-
How could I do it without
operator?
You are asking a lot of questions that you could answer yourself by reading the documentation, so I’ll give you a general advice: read it and experiment in the python shell. You’ll see that itemgetter returns a callable:
>>> func = operator.itemgetter(1)
>>> func(a)
['Paul', 22, 'Car Dealer']
>>> func(a[0])
8
To do it in a different way, you can use lambda:
a.sort(key=lambda x: x[1])
And reverse it:
a.sort(key=operator.itemgetter(1), reverse=True)
Sort by more than one column:
a.sort(key=operator.itemgetter(1,2))
See the sorting How To.
Looks like you’re a little bit confused about all that stuff.
operator is a built-in module providing a set of convenient operators. In two words operator.itemgetter(n) constructs a callable that assumes an iterable object (e.g. list, tuple, set) as input, and fetches the n-th element out of it.
So, you can’t use key=a[x][1] there, because python has no idea what x is. Instead, you could use a lambda function (elem is just a variable name, no magic there):
a.sort(key=lambda elem: elem[1])
Or just an ordinary function:
def get_second_elem(iterable):
return iterable[1]
a.sort(key=get_second_elem)
So, here’s an important note: in python functions are first-class citizens, so you can pass them to other functions as a parameter.
Other questions:
- Yes, you can reverse sort, just add
reverse=True:a.sort(key=..., reverse=True) - To sort by more than one column you can use
itemgetterwith multiple indices:operator.itemgetter(1,2), or with lambda:lambda elem: (elem[1], elem[2]). This way, iterables are constructed on the fly for each item in list, which are than compared against each other in lexicographic(?) order (first elements compared, if equal – second elements compared, etc) - You can fetch value at [3,2] using
a[2,1](indices are zero-based). Using operator… It’s possible, but not as clean as just indexing.
Refer to the documentation for details:
Answer for Python beginners
In simpler words:
- The
key=parameter ofsortrequires a key function (to be applied to be objects to be sorted) rather than a single key value and - that is just what
operator.itemgetter(1)will give you: A function that grabs the first item from a list-like object.
(More precisely those are callables, not functions, but that is a difference that can often be ignored.)
#sorting first by age then profession,you can change it in function "fun".
a = []
def fun(v):
return (v[1],v[2])
# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])
a.sort(key=fun)
print a
a = []
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])
print a
[['Nick', 30, 'Doctor'], ['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Mark', 66, 'Retired']]
def _cmp(a,b):
if a[1]<b[1]:
return -1
elif a[1]>b[1]:
return 1
else:
return 0
sorted(a,cmp=_cmp)
[['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Nick', 30, 'Doctor'], ['Mark', 66, 'Retired']]
def _key(list_ele):
return list_ele[1]
sorted(a,key=_key)
[['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Nick', 30, 'Doctor'], ['Mark', 66, 'Retired']]
>>>
The easiest way to sort an array using a user-defined function is to use cmp_to_key from Functools.
here is a sample code:
from functools import cmp_to_key
def mine(x,y):
if(x[1]!=y[1]): return x[1]>y[1]
else: return x[0]>y[0]
a = []
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 22, "Car Dealer"])
a.append(["Mark", 66, "Retired"])
def mine(a,b):
if a[1] > b[1]:
return 1
elif a[1] < b[1]:
return -1
else:
if a[0] > b[0]:
return 1
else:
return 0
print(sorted(a,key = cmp_to_key(mine)))