Accessing list values
Lists allow to use slice notation as lst[start:end:step] . The output of the slice notation is a new list containing elements from index start to end-1 . If options are omitted start defaults to beginning of list, end to end of list and step to 1:
lst = [1, 2, 3, 4] lst[1:] # [2, 3, 4] lst[:3] # [1, 2, 3] lst[::2] # [1, 3] lst[::-1] # [4, 3, 2, 1] lst[-1:0:-1]# [4, 3, 2] lst[5:8] # [] since starting index is greater than length of lst, returns empty list lst[1:10] # [2, 3, 4] same as omitting ending index
With this in mind, you can print a reversed version of the list by calling
lst[::-1] # [4, 3, 2, 1]
When using step lengths of negative amounts, the starting index has to be greater than the ending index otherwise the result will be an empty list.
lst[3:1:-1] # [4, 3]
# Using negative step indices are equivalent to the following code:
reversed(lst)[0:2] # 0 = 1 -1
# 2 = 3 -1
#The indices used are 1 less than those used in negative indexing and are reversed.Advanced slicing:
When lists are sliced the getitem() method of the list object is called, with a slice object. Python has a builtin slice method to generate slice objects. We can use this to store a slice and reuse it later like so,
data = 'chandan purohit 22 2000' # assuming data fields of fixed length name_slice = slice(0,19) age_slice = slice(19,21) salary_slice = slice(22,None) #now we can have more readable slices print(data[name_slice]) #chandan purohit print(data[age_slice]) #'22' print(data[salary_slice]) #'2000'
Iterating over a list:
my_list = ['foo', 'bar', 'baz']
for item in my_list:
print(item)
# Output: foo
# Output: bar
# Output: baz
for (index, item) in enumerate(my_list):
print('The item in position {} is: {}'.format(index, item))
# Output: The item in position 0 is: foo
# Output: The item in position 1 is: bar
# Output: The item in position 2 is: baz
Any and All
nums = [1, 1, 0, 1] all(nums) # False chars = ['a', 'b', 'c', 'd'] all(chars) # True
Likewise, any() determines if one or more values in an iterable evaluate to True
nums = [1, 1, 0, 1] any(nums) # True vals = [None, None, None, False] any(vals) # False
While this example uses a list, it is important to note these built-ins work with any Iterable, including generators.
vals = [1, 2, 3, 4] any(val > 12 for val in vals) # False any((val * 2) > 6 for val in vals) # True
Concatenate and Merge lists
# simplest way to concatenate list1 and list2 merged = list1 + list2
zip returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables:
alist = ['a1', 'a2', 'a3'] blist = ['b1', 'b2', 'b3'] for a, b in zip(alist, blist): print(a, b) # Output: # a1 b1 # a2 b2 # a3 b3
If the lists have different lengths then the result will include only as many elements as the shortest one:
alist = ['a1', 'a2', 'a3'] blist = ['b1', 'b2', 'b3', 'b4'] for a, b in zip(alist, blist): print(a, b) # Output: # a1 b1 # a2 b2 # a3 b3 alist = [] len(list(zip(alist, blist))) # Output: # 0
For padding lists of unequal length to the longest one with None s use itertools.zip_longest ( itertools.izip_longest in Python 2)
alist = ['a1', 'a2', 'a3'] blist = ['b1'] clist = ['c1', 'c2', 'c3', 'c4'] for a,b,c in itertools.zip_longest(alist, blist, clist): print(a, b, c) # Output: # a1 b1 c1 # a2 None c2 # a3 None c3 # None None c4
Remove duplicate values in list
Removing duplicate values in a list can be done by converting the list to a set that is an unordered collection of distinct objects). If a list data structure is needed, then the set can be converted back to a list using the function list() :
names = ["aixk", "duke", "edik", "tofp", "duke"] list(set(names)) # Output: ['duke', 'tofp', 'aixk', 'edik'] # Note that by converting a list to a set the original ordering is lost.
Accessing values in nested list
Starting with a three-dimensional list and Accessing items in the list:
alist = [[[1,2],[3,4]], [[5,6,7],[8,9,10], [12, 13, 14]]] print(alist[0][0][1]) #2 #Accesses second element in the first list in the first list print(alist[1][1][2]) #10 #Accesses the third element in the second list in the second list
Performing support operations:
alist[0][0].append(11) print(alist[0][0][2]) #11 #Appends 11 to the end of the first list in the first list
Using nested for loops to print the list:
for row in alist: #One way to loop through nested lists for col in row: print(col) #[1, 2, 11] #[3, 4] #[5, 6, 7] #[8, 9, 10] #[12, 13, 14]
Note that this operation can be used in a list comprehension or even as a generator to produce efficiencies, e.g.:
[col for row in alist for col in row] #[[1, 2, 11], [3, 4], [5, 6, 7], [8, 9, 10], [12, 13, 14]]
Not all items in the outer lists have to be lists themselves:
alist[1].insert(2, 15) #Inserts 15 into the third position in the second list
Another way to use nested for loops. The other way is better but I’ve needed to use this on occasion:
for row in range(len(alist)): #A less Pythonic way to loop through lists for col in range(len(alist[row])): print(alist[row][col]) #[1, 2, 11] #[3, 4] #[5, 6, 7] #[8, 9, 10] #15 #[12, 13, 14] #Using slices in nested list: print(alist[1][1:]) #[[8, 9, 10], 15, [12, 13, 14]] #Slices still work #The final list: print(alist) #[[[1, 2, 11], [3, 4]], [[5, 6, 7], [8, 9, 10], 15, [12, 13, 14]]]
Initializing a List to a Fixed Number of Elements
For immutable elements (e.g. None , string literals etc.):
my_list = [None] * 10 my_list = ['test'] * 10
For mutable elements, the same construct will result in all elements of the list referring to the same object, for example, for a set:
my_list=[{1}] * 10
print(my_list)
# [{1}, {1}, {1}, {1}, {1}, {1}, {1}, {1}, {1}, {1}]
my_list[0].add(2)
print(my_list)
# [{1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}]Instead, to initialize the list with a fixed number of different mutable objects, use:
my_list=[{1} for i in range(10)]
# my_list=[{1} for _ in range(10)]