Python - lambdas

lambdas.py

#-------------- Concept: Lambdas & string splits: --------------
TurnSpacesIntoColonsFunction = lambda x: '::'.join(x.split(" "))
NumberOfWordsFunction = lambda x: len(x.split(" "))

#-------------- Concept: lambdas inside a regex --------------

re.sub('[aeiou]', lambda match: match.group().upper()*3, 'abcdefghijklmnopqrstuvwxyz')


#-------------- Concept: crazy function stacking with lambdas --------------
# d(c(b("some4 string 56 with 3numb3rs in")))
# -> totals up the numbers

### strip the characters replace with ' ' turn into a list
b = lambda x: re.sub("\\D"," ", x).split(' ')

### filter out all the empty strings list elements
c = lambda y: filter(lambda x: x, y)

#### sum them
d = lambda list: reduce(lambda x,y: int(x)+int(y), list)

###### an easier way to do it:
reduce(lambda x,y : int(x) + int(y), re.findall("\\d","asdf23ds321"))

####### using reduce without lambdas:
def value_of_word():
  def count(i, a):
    return i + ord(a) - 64	
  return reduce(count, w, 0)

####### simple filters:
def isMp3(s):
    if s.find(".mp3") == -1:
        return False
    else:
        return True
 
list = ["1.mp3","2.txt", "3.mp3", "4.wmv","5.mp4" ]
temp = filter(isMp3,list)


####### 3 interesting ways of doing the same thing string -> objectId mapping
    return map((lambda x: ObjectId(x)), list)
    return map(ObjectId, list)
    return [ObjectId(x) for x in list]


######## Lists:
l = [1,2,3,4,5]

l[1:] = [2, 3, 4, 5]

l[2:] = [3, 4, 5]

l[-2:] = [4, 5]

l[::2] = [1, 3, 5] # Every 2nd value

l[::-2] = [5, 3, 1] # Every 2nd value backwards

# Rule: Python sequence slice addresses can be written as a[start:end:step] and any of start, stop or end can be dropped. a[::3] is every third element of the sequence.