Question: Lambda sort by second value of a list
>>> a = [[1,2], [0,1], [4,3], [4,5], [3,1]]
>>> a = sorted(a, key = lambda x:x[1])
>>> a
[[0, 1], [3, 1], [1, 2], [4, 3], [4, 5]]

Question: How to sort a dict by keys
mydict = {'carl':40, 'alan':2, 'bob':1, 'danny':3}
keylist = mydict.keys()
keylist.sort()
for key in keylist:
    print "%s: %s" % (key, mydict[key])
alan: 2
bob: 1
carl: 40
danny: 3

Question: Sort a dict by value
for key, value in sorted(mydict.iteritems(), key=lambda (k,v): (v,k)):
    print "%s: %s" % (key, value)
bob: 1
alan: 2
danny: 3
carl: 40

Question: ASCII <---> char
>>> ord('a') 
97
>>> chr(97) 
'a'
>>> chr(ord('a') + 3) 
'd'

Question: Reverse any list or string,  s
>>> s = "abc"     a = [1,2,3,4]
>>> s[::-1]       a= a[::-1]
'Cba'    
>>> s             a
'Abc'             [4,3,2,1]

Question: Reverse a list with reversed function:
>>> a = ["foo", "bar"] 
>>> for i in reversed(a): 
      print i 
Bar
Foo

WARNING! : altering string characters by index is not allowed. replace(“src”,”dst”) allowed.
To do that we have to convert s to a list then operate on it.
ret = list(s)
ret =  ret[::-1] #or, reversed(ret)  # ALT: use while to swap list items s[i] <--> s[n]
ret = "".join(ret)
return ret

Removing a key from dictionary  → a.pop(k, None)
my_dict.pop('key', None)


Question: Can we have integers as keys to Dictionaries? yes: 
>>> a = {1:10,2:"val2","key3":30}
>>> a
{'key3': 30, 1: 10, 2: 'val2'}
>>> a[1]
10
>>> a['key3']
30

Question: xrange-for (is faster than)  enumerate-for (is much faster than) pop-while
$ python -m timeit  'a = []' 'for i in xrange(10000): a.append(i)' 'for i in a: x=i'     →    1000 loops, best of 3: 1.06 msec per loop
$ python -m timeit  'a = []' 'for i in xrange(10000): a.append(i)' 'for index, val in enumerate(a):x=val'
1000 loops, best of 3: 1.27 msec per loop
$ python -m timeit  'a = []' 'for i in xrange(10000): a.append(i)' 'while a: x = a.pop(0)'       →   100 loops, best of 3: 15.4 msec per loop

Question: Python hash calculation
hash("ab")
>>> 12416074593111939
>>> hash(1)
>>> 1
>>> map(hash, (0, 1, 2, 3))
[0, 1, 2, 3]
>>> map(hash, ("namea", "nameb", "namea"))
[-1658398457, -1658398460, -1658398457]
>>> map(hash, ("ab", "ba", "cd", "ab"))
[12416074593111939, 12544075362113221, 12672076131114255, 12416074593111939]
https://www.laurentluce.com/posts/python-dictionary-implementation/

Question: Character to integer, lower() and discard special characters
>>> ord(‘A’) is 65            ord(‘Z’) = 90
>>> ord(‘a’) is 97            ord(‘z’) = 122
>>> for i, ch in enumerate("Word.Ok!".lower()):
...  if ch >='a' and ch <='z':
...    print ch
w
o
r
d
o
k
Collections/Counter, # occurrences of words in a document or,  list of words
>>> from collections import Counter   # remember Counter not count
>>> cnt = Counter()
>>> for word in ['red', 'blue', 'red', 'green', 'blue', 'blue']:
...    cnt[word] += 1
>>> cnt
Counter({'blue': 3, 'red': 2, 'green': 1})

Question: >>> # Find the ten most common words in Hamlet
>>> import re
>>> words = re.findall(r'\w+', open('hamlet.txt').read().lower())
>>> Counter(words).most_common(10)
[('the', 1143), ('and', 966), ('to', 762), ('of', 669), ('i', 631),
 ('you', 554),  ('a', 546), ('my', 514), ('hamlet', 471), ('in', 451)]

>>> mydict = {'carl':40, 'alan':2, 'bob':1, 'danny':3,'don':3}
    >>>
>>> sorted(mydict.iteritems(), key=lambda (k,v): (v,k)) 
[('bob', 1), ('alan', 2), ('danny', 3), ('don', 3), ('carl', 40)]

You only need to return the value that you are comparing in the lambda, as opposed to flipping the key and value in a tuple

>>> sorted(mydict.iteritems(), key=lambda (k,v): v)
[('bob', 1), ('alan', 2), ('danny', 3), ('don', 3), ('carl', 40)]

Checking if a object is an instance of a class:
if isinstance(obj, MyClass):
     print "obj is my object"

Question: Data type conversion
→ reversing a integer number in 32 bit and return back
n = 14
b = bin(n)  #binary of ‘a’ -- “0b1110”
# also have hex, oct for hexadecimal and octal
#OR b= b.replace(“0b”,“”)
dummy = [“0” for i in xrange(32-len(b) )]
b = list(b)  # make list
b  = dummy + b
b = b[::-1]  # reverse
b = “”.join(b)
#convert back to integer and return
return int(b,2) #if there is 0x--hex or 0b--bin we can just replace it for operation

Large Numbers
If you just need a number that's bigger than all others, you can use
>> float('inf')

in similar fashion, a number smaller than all others:
>> float('-inf')

The value sys.maxint, which is either 231 - 1 or 263 - 1 depending on your platform.

→ googol
>>> 10**100
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000L

>>> a = 10**100
>>> b = sys.maxint
>>> if a >b:
...   print "googol is beggier than maxint"
...
... googol is beggier than maxint

>>> long(1234567891011121314151617181920)
>>> 1234567891011121314151617181920L

>>> a = sys.maxint
>>> b = float("inf")
>>> if b>a:
...   print "inf is bigger"
... else:
...   print "maxint is bigger"
...
... inf is bigger


Randoms
import random

>>> random.random()  # gives a random  0<=n<=1
0.010835814926994924

>>> random.randint(a,b)  # gives a random  a<=n<=b

* If a seed is set then you will get the same result right after setting the seed
random.seed(a)  # a is integer


Question: Bitwise operations
a = 60            # 60 = 0011 1100 
b = 13            # 13 = 0000 1101 
c = 0

c = a & b;        # 12 = 0000 1100    AND
c = a | b;        # 61 = 0011 1101    OR
c = a ^ b;        # 49 = 0011 0001    EX-OR
c = ~a;           # -61 = 1100 0011    NOT
c = a << 2;       # 240 = 1111 0000    Shift left twice, with 0 from right
c = a >> 2;       # 15 = 0000 1111    Shift Right twice, with 0 from left
Singly linked list palindrome: 
Choice 1: make a string list and push by traverse then match .
Choice 2: Solve in O(n) time and O(1) extra space.
1) Get the middle of the linked list by getting length and going to mid node. 
2) make a new pointer node after the middle (even/odd index matters)
4) reverse second half.
3) traverse both the list. Check if the first half and second half are identical.
------
4) Construct the original linked list by reversing the second half again and attaching it back to the first half

Copy a list in python--
a =[1,2,3]
b= a[:]
#OR
b= list(a)

Set operations in python:
myset.add(data)
myset.delete(data)
x = myset.pop()  # can only pop first element or raise exception

Distance between two points,  d =  (x1-x2)2 + (y1-y2)2 
Area of a triangle:  ½ *length*height
import math
S = (a+b+c) / 2.0  #half of the triangles perimeter
#Area =  S*(S-a)*(S-b)*(S-c) 
area = math.sqrt (s*(s-a)*(s-b)*(s-c))

Two points  (x1, y1) and (x2, y2), slope is   y2- y1x2- x1
Roman numerals: I=1; X =10, V=5 L=50, C=100, D =500, M=1000
http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm
roman to integer.
    a) convert str to a list. b) convert each character to integer equivalent
            c*) if digits[i] >digits[i-1]: digits[i] =-1*digits[i]       d)return sum(digits)

Python Strings
Creation
the_string = “Hello World!”
The_string[4]
#returns ‘o’


Split
the_string.split(‘ ’)   → nothing in ‘’ means space ‘ ’
[‘Hello’, ‘World!’] 

Replace words/substrings
str2 = the_string.replace(‘Hello’, “OK”)
“OK World!”

Join
words = [“this”, ‘is’, ‘a’, ‘list’, ‘of’, “strings”]
ret = ‘’.join(words)
#This is a list of strings
‘ZOOL’.join(words)
#ThisZOOLisZOOLaZOOLlistZOOLofZOOLstrings
‘’.join(words)
#Thisisalistofstrings

Print "codecodecodecode mentormentormentormentormentor" without using loops
print "code"*4+' '+"mentor"*5

To count occurance of a characeter, use  count(), returns 
if s.count('(')!=s.count(')') or s.count('[')!=s.count(']')




Counter Method
from collections import Counter
def is_anagram(str1, str2):
     return Counter(str1) == Counter(str2)
>>> is_anagram('abcd','dbca')
True
>>> is_anagram('abcd','dbaa')
False

Substring find:         str.find(what, beg=0, end=len(string))
what − search string beg − beginning index, [default 0, end − ending[default=n-1]
Return : - index if found and   -1  otherwise.

str1 = "this is string example....wow!!!";
str2 = "exam";
print str1.find(str2)
print str1.find(str2, 10)
print str1.find(str2, 40)
15
15
-1

Find all occurances of a substring
import re
[m.start() for m in re.finditer('test', 'test test test test')] → iteratively what, where
#[0, 5, 10, 15]
>>> [[m.start(),m.end()] for m in re.finditer('test', 'test test test')]
>>> [[0, 4], [5, 9], [10, 14]]

If you want to find overlapping matches, lookahead will do that:
[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]

Python Tuples
A tuple consists of a number of values separated by commas. They are useful for ordered pairs and returning several values from a function.
Creation: 
emptyTuple = () 
singleItemTuple = (“spam”,)
thistuple = 12, 89, 'a'
thistuple = (12, 89, 'a')               #note the comma

Accessing: 
thistuple[0]    →  returns 12

Python Dictionaries
A dictionary is a set of key:value pairs. All keys in a dictionary must be unique.
Creation:
emptyDict = {}
    mydict = {'a':1, 'b':23, 'c':"eggs"}

Accessing: deleting: finding:
mydict['a']        →  returns 1
del mydict['b']
mydict.has_key('e')  → returns True/False
    
>>> mydict = {'a':1, 'b':2, 3:'c'}
>>> mydict.keys()
['a', 3, 'b']
>>> mydict.values()
[1, 'c', 2]
>>> mydict.items()
[('a', 1), (3, 'c'), ('b', 2)]

Python dictionaries implement a tp_iter slot that returns an efficient iterator of keys.
for k in mydict: …    #equivalent to, but much faster than      for k in mydict.keys(): …
    Example: ret = [key for key in mydict if mydict[key]==True] [0]

Iterate key, value pairs:
i.       for key in dict.iterkeys():   #for x in dict is shorthand for for x in dict.iterkeys().
ii.  for value in dict.itervalues():
iii. for key, value in dict.iteritems(): …
iv.  for item in dict.items():   #python3 and faster than iter items



--> List : one of the most important data structures in Python is the list. Lists are very flexible and have many built-in control functions.

>>> [0]*8
[0, 0, 0, 0, 0, 0, 0, 0]
>>> ['ok'] * 4
['ok', 'ok', 'ok', 'ok']

mylist.count()
mylist = [123, 'xyz', 'zara', 'abc', 123]
print mylist.count(123), mylist.count(99)
-----------
2 0


creation: accessing: slicing:
length: sort: add: return &
remove: insert: remove:
thelist = [5,3,‘p’,9,‘e’]


[5,3,’p’,9,’e’]
[5,3,’p’,9,’e’]
[5,3,’p’,9,’e’]
[5,3,’p’,9,’e’]
[5,3,’p’,9,’e’]
[5,3,’p’,9,’e’]
[5,3,’p’,9,’e’]
thelist.sort()  → no return value

thelist.append(37)
thelist.pop() returns 37
thelist.pop(1) returns 5
thelist.insert(2, ‘z’)
thelist.remove(‘e’)
del thelist[0]


[3,5,9,’e’,’p’]
[3,5,9,’e’,’p’,37]
[3,5,9,’e’,’p’]
[3,9,’e’,’p’]
[3,’z’,9,’e’,’p’]
[3,’z’,9,’p’]
[‘z’,9,’p’]
[‘z’,9,’p’]
[‘z’,9,’p’]


‘c’ in thisdict
‘paradimethylaminobenzaldehyde’ in thisdict returns False
thelist[0]
Thelist[1:3]   similar to [thelist[i] for i in xrange(1,3)]
thelist[2:]
thelist[:2]
thelist[2:-1]
len(thelist)


returns 5 returns [3, ‘p’]
returns [‘p’, 9, ‘e’] returns [5, 3]
returns [‘p’, 9] returns 5
thelist.sort()
thelist.append(37)


no return value
thelist.pop() returns 37thelist.pop(1) returns 5thelist.insert(2, ‘z’) thelist.remove(‘e’)
del thelist[0]concatenation: thelist + [0]
finding: 9 in thelist
returns [‘z’,9,’p’,0] returns True

returns False
returns [‘a’, ‘c’]
returns [(‘a’, 1), (‘c’, ‘eggs’)] 
returns True

List Comprehension: A special expression enclosed in square brackets that returns a new list. The expression is of the form:[expression for expr in sequence if condition] The condition is optional.

A = [d for d in xrange(5) if d%2 == 0]



##### Files
Open:
thisfile = open(“datadirectory/file.txt”, “r”)  # r, w → read,write

Accessing:
infile.read()        → reads entire file into one string 
infile.readline()        → reads one line of a file
lines = infile.readlines()        → reads entire file into a list of strings, one per line
for eachline in infile:    → steps through lines in a file
    print eachline
import os.path 
os.path.isfile(fname)  → returns True or False