- Python Basics
- Collections
- Control Flow
- Functions
- Classes
- Modules
- Generators and Decorators
- Python Next
Example code based on
LearnXinYminutes
and licensed under
CC Attribution-Share 3
Collections
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Lists
# Lists store sequences (like an Array in Ruby)
li = []
# You can start with a prefilled list
other_li = [4, 5, 6]
# Add stuff to the end of a list with append
li.append(1) #li is now [1]
li.append(2) #li is now [1, 2]
li.append(4) #li is now [1, 2, 4]
li.append(3) #li is now [1, 2, 4, 3]
# Remove from the end with pop
li.pop() #=> 3 and li is now [1, 2, 4]
# Let's put it back
li.append(3) # li is now [1, 2, 4, 3] again.
# Access a list like you would any array
li[0] #=> 1
# Look at the last element
li[-1] #=> 3
# Looking out of bounds is an IndexError
li[4] # Raises an IndexError
# You can look at ranges with slice syntax.
# (It's a closed/open range for you mathy types.)
li[1:3] #=> [2, 4]
# Omit the beginning
li[2:] #=> [4, 3]
# Omit the end
li[:3] #=> [1, 2, 4]
# Select every second entry
li[::2] #=>[1,4]
# Revert the list
li[::-1] #=> [3, 4, 2, 1]
# Use any combination of these to make advanced slices
# li[start:end:step]
# Remove arbitrary elements from a list with "del"
del li[2] # li is now [1, 2, 3]
# You can add lists
li + other_li #=> [1, 2, 3, 4, 5, 6] - Note: li and other_li is left alone
# Concatenate lists with "extend()"
li.extend(other_li) # Now li is [1, 2, 3, 4, 5, 6]
# Check for existence in a list with "in"
1 in li #=> True
# Examine the length with "len()"
len(li) #=> 6
Tuples
# Tuples are like lists but are immutable.
tup = (1, 2, 3)
tup[0] #=> 1
tup[0] = 3 # Raises a TypeError
# You can do all those list thingies on tuples too
len(tup) #=> 3
tup + (4, 5, 6) #=> (1, 2, 3, 4, 5, 6)
tup[:2] #=> (1, 2)
2 in tup #=> True
# You can unpack tuples (or lists) into variables
a, b, c = (1, 2, 3) # a is now 1, b is now 2 and c is now 3
# Tuples are created by default if you leave out the parentheses
d, e, f = 4, 5, 6
# Now look how easy it is to swap two values
e, d = d, e # d is now 5 and e is now 4
Dictionaries
# Dictionaries store key-value mappings (like Hashes in Ruby)
empty_dict = {}
# Here is a prefilled dictionary
filled_dict = {"one": 1, "two": 2, "three": 3}
# Look up values with []
filled_dict["one"] #=> 1
# Get all keys as a list with "keys()"
filled_dict.keys() #=> ["three", "two", "one"]
# Note - Dictionary key ordering is not guaranteed.
# Your results might not match this exactly.
# Get all values as a list with "values()"
filled_dict.values() #=> [3, 2, 1]
# Note - Same as above regarding key ordering.
# Check for existence of keys in a dictionary with "in"
"one" in filled_dict #=> True
1 in filled_dict #=> False
# Looking up a non-existing key is a KeyError
filled_dict["four"] # KeyError
# Use "get()" method to avoid the KeyError
filled_dict.get("one") #=> 1
filled_dict.get("four") #=> None
# The get method supports a default argument when the value is missing
filled_dict.get("one", 4) #=> 1
filled_dict.get("four", 4) #=> 4
# "setdefault()" inserts into a dictionary only if the given key isn't present
filled_dict.setdefault("five", 5) #filled_dict["five"] is set to 5
filled_dict.setdefault("five", 6) #filled_dict["five"] is still 5
Sets
# Sets store ... well sets
empty_set = set()
# Initialize a "set()" with a bunch of values
some_set = set([1,2,2,3,4]) # some_set is now set([1, 2, 3, 4])
# Since Python 2.7, {} can be used to declare a set
filled_set = {1, 2, 2, 3, 4} # => {1, 2, 3, 4}
# Add more items to a set
filled_set.add(5) # filled_set is now {1, 2, 3, 4, 5}
# Do set intersection with &
other_set = {3, 4, 5, 6}
filled_set & other_set #=> {3, 4, 5}
# Do set union with |
filled_set | other_set #=> {1, 2, 3, 4, 5, 6}
# Do set difference with -
{1,2,3,4} - {2,3,5} #=> {1, 4}
# Check for existence in a set with in
2 in filled_set #=> True
10 in filled_set #=> False
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Challenge
You will be given an array (or "list") of numbers ar
. Create and print another array of length 2 that just contains the first and last number in the original array.
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Comments
Sirou Dimicom
Feb 2, 8:14 AMWhat is the answer please of this challenge
Learneroo
Feb 2, 8:23 AMAnswer added.