FirstJupiterNotebook.py

#!/usr/bin/env python
# coding: utf-8

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# LIST  --- collection of certain items, separated by coma between square brackets


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L = [1, '11', "1da", [1, 3], ["X", 10, (11, 7)]] # lists can have different types of elements


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type(L) # return the type of a variable


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empty = [] # empty list has no elements
empty


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len(empty) # number of elements in a list


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f = [11, 12, 13, 14] # create a list of elements 11, 12, 13, 14
f


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f += [15, 16] # add 15 and 16 right after 14 to f
f


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# to add a single element 17 to the end of the list f
f.append(17) #, more likely append is the same as the incrementation, but takes exactly 1 element.
f


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# f.append(18, 19, 20) # is not correct


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f.extend([18, 19, 20])
f


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# insert element at a particular location
f.insert(0,8) # insert 8 to the first element (which is indexed by 0)
f


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f.insert(1,9)
f


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f.index(100) # returns the first index of a value 9 in a list f if the value already exists in a list


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9 in f #decide simply if an element appears in a list


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#if an element appears in a list we can return its index


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b = -1
if 100 in f:
    b = f.index(100)
b


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b = -1
if 9 in f:
    b = f.index(9)
b


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# if an element exists in a list we can insert something directly after that.
f.insert(f.index(9)+1, 10)
f


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f.remove(8) # remove element 8 from list f
f


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f.remove(8) # remove element 8 from list f (only if the element exists in the list)
f


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p = f.pop() # remove and return the last element from the list
f


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p


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f.sort() # sort elements (by default in ascending order)
f


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f.sort(reverse=True) # sort elements in descending order
f


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f.reverse() # reverse the list
f


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max(f)


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min(f)


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f.count(10) # how many times an item appears in a list


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f.count(5)


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f.clear() # remove all the elements from a list
f == []


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L = list(range(10))
L


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L[0] # get the first element from the list


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L[1] # get the second element from the list


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len(L) # number of elements in the list, last element is indexed by len(L)-1


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L[len(L)-1] # the last element in the list


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L[len(L)-2] # second element from the back


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L[-1] # different way of indexing elements last element is indexed by -1


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L[-2] == L[len(L)-2]


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list(range(10)) # list of numbers from 0 to 10 (including 0, but excluding 10, range(10)=[0,10))


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g = []
k = 0
while k < 10:
    if (k % 2 == 0):
        g.append(k)
    k+=1
del k
g


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g = []
for k in range(10):
    if (k % 2 == 0):
        g.append(k)
g


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[k for k in range(10) if k % 2 == 0] # create a list of even numbers using list comprehension


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[L[0], L[1], L[2]] # create a new list of the first three element from the list


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[L[k] for k in range(3)]


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L[0:3] # first 3 elements in the list L[0:3] = [L[0], L[1], L[2]]


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L[:3] # if the first argument is not specified, it means it is zero (L[:3]==L[0:3])


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L[3:] # if the second argument is not specified it means it is the last element


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L[:] # from the first to the last (both arguments neither specified) which means it is the whole list


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list(range(0,10,2)) # exactly the same, but the most efficient way (create a range from 0 to 10 but exclude every second item)


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L[1:4] # 3 elements after the first


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[L[4], L[5], L[6]]


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[L[k] for k in range(4,7)]


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L[4:7]


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L[4:7:1] # specify a skipping (1 means every element included between 4th and 7th (7th still excluded))


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L[4:7:2] # hold every second element


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L[-3:-1] # -1 is the last element (which is excluded) 


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L[-3:] # we can include the last element as well


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L[-3:-1:1] # number of steps can be specified


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L[:-4:-1] # if the number of steps is negative it means the order of the elements will be reversed


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L[::-1] # all the elements in reversed order


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L.reverse()
L


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reversed(L)


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list(reversed(L))


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L[1:3] = [11, 12] # slice assignment
L


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L[1:3]=[] # remove elements using slicing assignment
L


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LL = L # looks like this assignment create a copy of list L, but it is just a reference
LL


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LL[0]=100 # we change the first element of LL
LL


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L # as LL is a reference to L, L[0] changed to 100 as well


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LLL=list(LL) # now we copied the LL list, LLL is not a reference to LL
LLL


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LLL[0]=1
LLL


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LL # LL list remains the same


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# this approach fails when the list has list elements


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X = [[10, 2], [20, 3], [30, 4]]
X


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Y = list(X)
Y


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Y[0][0] = 100
Y


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X


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import copy


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Y = copy.deepcopy(X)


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X[0][0] = 10
X


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Y


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import itertools


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flatY=list(itertools.chain(*Y))


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flatY


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X = [1, 1, 2, 3, 3, 4, 5, 3, 2, 3]
X


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[k for k in X if X.count(k)>1] # find all the repeated elements


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list(set([k for k in X if X.count(k)>1])) # these are the duplicated elements in X


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from collections import deque


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Z=deque(X)


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Z.rotate(1)


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list(Z)


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import numpy as np


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list(np.arange(0,10, 2))


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np.linspace(0, 8, 5, dtype="int")  # by default, np.linspace returns an array filled with doubles


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#  STRING


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sentence = "I got this feeling on the summer day when you were gone"
sentence


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type(sentence)


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sentence[0] # we can access the characters in the string the same way as in lists


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sentence[-1]


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sentence_list = sentence.split(" ")
sentence_list


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type(sentence_list)


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s = " ".join(sentence_list)
s


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word = "hello"


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word_list = list(word) # create a list from a string


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integer = 12345
integer


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integer_list = list(integer) # this not works


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integer_list = list(str(integer)) # is there any way to specify datatype ???
integer_list


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np.array(integer, dtype="int")


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[k for k in str(integer)]


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# DICTIONARY


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d={'A': 1, 'B' : 2, 'C' : 3, 'D' : 4}


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d


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d.keys()


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type(d.keys())


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d.values()


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type(d.values())


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