First attempts
OMG! My first Python code works! (P.S. Week ago I didn't know Python exists)
import string
punct = set(string.punctuation + string.digits)
fin_sentence = ['.','...','?','!','?!']
longest = None
shortest = None
text = raw_input('Enter some text:')
free_text = ''.join(x for x in text if x not in punct) # deleting all digits and punctuation marks
punctuation = ''.join(x for x in text if x in set(string.punctuation))
sentences = ''.join(x for x in text if x in fin_sentence)
free_text = free_text.lower()
words = free_text.split() #making list named words
number = len(words)
total = 0
length = 0
counts = {} #dictionary
pop_lst = []
longest_lst = []
shortest_lst = []
for word in words:
total += len(word) #counting number of letters
if longest is None or len(word) > len(longest): #searching for longest word
longest = word
l = len(longest)
if shortest is None or len(word) < len(shortest): #searching for shortest word
shortest = word
s = len(shortest)
counts[word] = counts.get(word,0) + 1 #filling dictionary
v = counts.values()
for word in words:
if len(word) == l:
longest_lst.append(word)
if len(word) == s:
shortest_lst.append(word)
for word,count in counts.items():
if count == max(v):
pop_lst.append(word)
number_pop = len(pop_lst)
for word in pop_lst:
length += len(word)
if number < 1: #for no text entered
print 'number of words:', number
print 'end of program'
else:
print 'number of letters:', total
print 'number of words:', number
print 'number of unique words:', len(counts.keys())
print 'number of punctuation marks:', len(punctuation)
print 'number of sentences:', len(sentences)
print 'average word length is:', float(total)/number
print 'longest words are:', longest_lst, ',', 'length:', len(longest), ',', 'number of words in text:', len(longest_lst)
print 'shortest words are:', shortest_lst, ',', 'length:', len(shortest), ',', 'number of words in text:', len(shortest_lst)
print 'most popular words are', pop_lst, ',', 'repetitions:', max(v), 'average length is:', float(length)/number_pop
print 'end of program'