List methods and supported operators
suggest changeStarting with a given list a
:
a = [1, 2, 3, 4, 5]
append(value)
– appends a new element to the end of the list.# Append values 6, 7, and 7 to the list a.append(6) a.append(7) a.append(7) # a: [1, 2, 3, 4, 5, 6, 7, 7] # Append another list b = [8, 9] a.append(b) # a: [1, 2, 3, 4, 5, 6, 7, 7, [8, 9]] # Append an element of a different type, as list elements do not need to have the same type my_string = "hello world" a.append(my_string) # a: [1, 2, 3, 4, 5, 6, 7, 7, [8, 9], "hello world"] **Note that** the `append()` method only appends one new element to the end of the list. If you append a list to another list, the list that you append becomes a single element at the end of the first list. # Appending a list to another list a = [1, 2, 3, 4, 5, 6, 7, 7] b = [8, 9] a.append(b) # a: [1, 2, 3, 4, 5, 6, 7, 7, [8, 9]] a[8] # Returns: [8,9]
extend(enumerable)
– extends the list by appending elements from another enumerable.a = [1, 2, 3, 4, 5, 6, 7, 7] b = [8, 9, 10] # Extend list by appending all elements from b a.extend(b) # a: [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10] # Extend list with elements from a non-list enumerable: a.extend(range(3)) # a: [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 0, 1, 2] Lists can also be concatenated with the `+` operator. Note that this does not modify any of the original lists: a = [1, 2, 3, 4, 5, 6] + [7, 7] + b # a: [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10]
index(value, [startIndex])
– gets the index of the first occurrence of the input value. If the input value is not in the list aValueError
exception is raised. If a second argument is provided, the search is started at that specified index.a.index(7) # Returns: 6 a.index(49) # ValueError, because 49 is not in a. a.index(7, 7) # Returns: 7 a.index(7, 8) # ValueError, because there is no 7 starting at index 8
insert(index, value)
– insertsvalue
just before the specifiedindex
. Thus after the insertion the new element occupies positionindex
.a.insert(0, 0) # insert 0 at position 0 a.insert(2, 5) # insert 5 at position 2 # a: [0, 1, 5, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10]
pop([index])
– removes and returns the item atindex
. With no argument it removes and returns the last element of the list.a.pop(2) # Returns: 5 # a: [0, 1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10] a.pop(8) # Returns: 7 # a: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] # With no argument: a.pop() # Returns: 10 # a: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
remove(value)
– removes the first occurrence of the specified value. If the provided value cannot be found, aValueError
is raised.a.remove(0) a.remove(9) # a: [1, 2, 3, 4, 5, 6, 7, 8] a.remove(10) # ValueError, because 10 is not in a
reverse()
– reverses the list in-place and returnsNone
.a.reverse() # a: [8, 7, 6, 5, 4, 3, 2, 1] There are also [other ways of reversing a list][2].
count(value)
– counts the number of occurrences of some value in the list.a.count(7) # Returns: 2
sort()
– sorts the list in numerical and lexicographical order and returnsNone
.a.sort() # a = [1, 2, 3, 4, 5, 6, 7, 8] # Sorts the list in numerical order
Lists can also be reversed when sorted using the `reverse=True` flag in the `sort()` method.
a.sort(reverse=True) # a = [8, 7, 6, 5, 4, 3, 2, 1]
If you want to sort by attributes of items, you can use the `key` keyword argument:
import datetime class Person(object): def __init__(self, name, birthday, height): self.name = name self.birthday = birthday self.height = height def __repr__(self): return self.name l = [Person("John Cena", datetime.date(1992, 9, 12), 175), Person("Chuck Norris", datetime.date(1990, 8, 28), 180), Person("Jon Skeet", datetime.date(1991, 7, 6), 185)] l.sort(key=lambda item: item.name) # l: [Chuck Norris, John Cena, Jon Skeet] l.sort(key=lambda item: item.birthday) # l: [Chuck Norris, Jon Skeet, John Cena] l.sort(key=lambda item: item.height) # l: [John Cena, Chuck Norris, Jon Skeet]
In case of list of dicts the concept is the same:
import datetime l = [{'name':'John Cena', 'birthday': datetime.date(1992, 9, 12),'height': 175}, {'name': 'Chuck Norris', 'birthday': datetime.date(1990, 8, 28),'height': 180}, {'name': 'Jon Skeet', 'birthday': datetime.date(1991, 7, 6), 'height': 185}] l.sort(key=lambda item: item['name']) # l: [Chuck Norris, John Cena, Jon Skeet] l.sort(key=lambda item: item['birthday']) # l: [Chuck Norris, Jon Skeet, John Cena] l.sort(key=lambda item: item['height']) # l: [John Cena, Chuck Norris, Jon Skeet]
Sort by sub dict :
import datetime l = [{'name':'John Cena', 'birthday': datetime.date(1992, 9, 12),'size': {'height': 175, 'weight': 100}}, {'name': 'Chuck Norris', 'birthday': datetime.date(1990, 8, 28),'size' : {'height': 180, 'weight': 90}}, {'name': 'Jon Skeet', 'birthday': datetime.date(1991, 7, 6), 'size': {'height': 185, 'weight': 110}}] l.sort(key=lambda item: item['size']['height']) # l: [John Cena, Chuck Norris, Jon Skeet]
Better way to sort using
attrgetter
anditemgetter
Lists can also be sorted using
attrgetter
anditemgetter
functions from the operator module. These can help improve readability and reusability. Here are some examples,from operator import itemgetter,attrgetter people = [{'name':'chandan','age':20,'salary':2000}, {'name':'chetan','age':18,'salary':5000}, {'name':'guru','age':30,'salary':3000}] by_age = itemgetter('age') by_salary = itemgetter('salary') people.sort(key=by_age) #in-place sorting by age people.sort(key=by_salary) #in-place sorting by salary
itemgetter
can also be given an index. This is helpful if you want to sort based on indices of a tuple.list_of_tuples = [(1,2), (3,4), (5,0)] list_of_tuples.sort(key=itemgetter(1)) print(list_of_tuples) #[(5, 0), (1, 2), (3, 4)]
Use the
attrgetter
if you want to sort by attributes of an object,persons = [Person("John Cena", datetime.date(1992, 9, 12), 175), Person("Chuck Norris", datetime.date(1990, 8, 28), 180), Person("Jon Skeet", datetime.date(1991, 7, 6), 185)] #reusing Person class from above example person.sort(key=attrgetter('name')) #sort by name by_birthday = attrgetter('birthday') person.sort(key=by_birthday) #sort by birthday
clear()
– removes all items from the lista.clear() # a = []
- Replication – multiplying an existing list by an integer will produce a larger list consisting of that many copies of the original. This can be useful for example for list initialization:
b = ["blah"] * 3 # b = ["blah", "blah", "blah"] b = [1, 3, 5] * 5 # [1, 3, 5, 1, 3, 5, 1, 3, 5, 1, 3, 5, 1, 3, 5]
Take care doing this if your list contains references to objects (eg a list of lists), see [Common Pitfalls - List multiplication and common references][2].
- Element deletion – it is possible to delete multiple elements in the list using the
del
keyword and slice notation:a = list(range(10)) del a[::2] # a = [1, 3, 5, 7, 9] del a[-1] # a = [1, 3, 5, 7] del a[:] # a = []
- Copying
The default assignment "=" assigns a reference of the original list to the new name. That is, the original name and new name are both pointing to the same list object. Changes made through any of them will be reflected in another. This is often not what you intended.
b = a a.append(6) # b: [1, 2, 3, 4, 5, 6]
If you want to create a copy of the list you have below options.
You can slice it:
new_list = old_list[:]
You can use the built in list() function:
new_list = list(old_list)
You can use generic copy.copy():
import copy new_list = copy.copy(old_list) #inserts references to the objects found in the original.
This is a little slower than list() because it has to find out the datatype of old_list first. If the list contains objects and you want to copy them as well, use generic copy.deepcopy():
import copy new_list = copy.deepcopy(old_list) #inserts copies of the objects found in the original.
Obviously the slowest and most memory-needing method, but sometimes unavoidable.
copy()
– Returns a shallow copy of the listaa = a.copy() # aa = [1, 2, 3, 4, 5]