Difference between revisions of "Object and Data Structure Basics"
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newlist.sort() # sort list in place, which means you have to call newlist again to get the sorted results. you cannot do x = newlist.sort() becasue it will return nothing | newlist.sort() # sort list in place, which means you have to call newlist again to get the sorted results. you cannot do x = newlist.sort() becasue it will return nothing | ||
numlist.reverse() # reverse in place | numlist.reverse() # reverse in place | ||
==Tuples== | |||
Tuples are immutable | |||
t=(1,2,3) | |||
len(t) | |||
t[0] | |||
t.count(1) # will count the amount of time 1 appears | |||
t.index(1) # will return the first index that 1 occors | |||
==Dictionaries== | ==Dictionaries== | ||
'''Has key value pairs''' | '''Has key value pairs''' |
Revision as of 21:38, 3 January 2019
List
mylist= ['string', 1,2,3] len(mylist) # get length of list Result 3
Get part of a list (Slicing)
mylist[1:] gets from index one to the end result 1,2,3
Concatenate list
another_list = [5,6] newlist = mylist + anohter_list result ['string', 1,2,3,5,6]
Add an item to the end of a list
newlist.append('hello')
Remove item from list
newlist.pop() # will pop last item newlist.pop(0) # will pop index item 0
Sort and Reverse
newlist.sort() # sort list in place, which means you have to call newlist again to get the sorted results. you cannot do x = newlist.sort() becasue it will return nothing numlist.reverse() # reverse in place
Tuples
Tuples are immutable
t=(1,2,3) len(t) t[0] t.count(1) # will count the amount of time 1 appears t.index(1) # will return the first index that 1 occors
Dictionaries
Has key value pairs
mydict = {'key1':'value1', 'key2':'value2'} mydict[key1] out: value1 mydict2 = {'k1':123, 'k2':[0,1,2], 'k3':{'insidekey':100}} mydict2['k2'][2] mydict2['k3']['insidekey']
Add to dictonary
mydict2['k4'] = 300
List all keys in dictionary
mydict2.keys()
List all values in distionary
dict2.values()
List both items and keys
dict2.items()