In [13]:
count = 1
for elem in range(1, 3 + 1):
count *= elem
print(count)
In [14]:
from math import factorial as f
f(3)
Out[14]:
.1 + .1 + .1 == .3format(.1, '.100g')(0, 1, 2000000) < (0, 3, 4)Ex.
txt = 'but soft what light in yonder window breaks'
words = txt.split()
t = list()
for word in words:
t.append((len(word), word))
t.sort(reverse=True)
res = list()
for length, word in t:
res.append(word)
print(res)Why would words.sort() not work?
addr = 'monty@python.org'
uname, domain = addr.split('@')t = (a, b, c)
func(*t)Revise a previous program as follows: Read and parse the "From" lines and pull out the addresses from the line. Count the number of messages from each person using a dictionary.
After all the data has been read print the person with the most commits by creating a list of (count, email) tuples from the dictionary and then sorting the list in reverse order and print out the person who has the most commits.
Sample Line:
From stephen.marquard@uct.ac.za Sat Jan 5 09:14:16 2008
Enter a file name: mbox-short.txt
cwen@iupui.edu 5
Enter a file name: mbox.txt
zqian@umich.edu 195This program counts the distribution of the hour of the day for each of the messages. You can pull the hour from the "From" line by finding the time string and then splitting that string into parts using the colon character. Once you have accumulated the counts for each hour, print out the counts, one per line, sorted by hour as shown below.
Sample Execution:
python timeofday.py
Enter a file name: mbox-short.txt
04 3
06 1
07 1
09 2
10 3
11 6
14 1
15 2
16 4
17 2
18 1
19 1Write a program that reads a file and prints the letters in decreasing order of frequency. Your program should convert all the input to lower case and only count the letters a-z. Your program should not count spaces, digits, punctuation or anything other than the letters a-z. Find text samples from several different languages and see how letter frequency varies between languages. Compare your results with the tables at wikipedia.org/wiki/Letter_frequencies.
In [40]:
def n_max():
inpt = eval(input("Please enter some values: "))
maximum = max_val(inpt)
print("The largest value is", maximum)
def max_val(ints):
"""Input: collection of ints.
Returns: maximum of the collection
int - the max integer."""
max = ints[0]
for x in ints:
if x > max:
max = x
return max
assert max_val([1, 2, 3]) == 3
assert max_val([1, 1, 1]) == 1
assert max_val([1, 2, 2]) == 2
In [41]:
n_max()
In [31]:
inpt = eval(input("Please enter three values: "))
In [34]:
list(inpt)
Out[34]:
Strategy 1: Compare each to all (brute force)
Strategy 2: Decision Tree
Strategy 3: Sequential processing
Strategy 4: Use python
In [ ]:
assert compress('AAAADDBBBBBCCEAA') == 'A4D2B5C2E1A2'
In [68]:
# %load ../scripts/compress/compressor.py
def groupby_char(lst):
"""Returns a list of strings containing identical characters.
Takes a list of characters produced by running split on a string.
Groups runs (in order sequences) of identical characters into string elements in the list.
Parameters
---------
Input:
lst: list
A list of single character strings.
Output:
grouped: list
A list of strings containing grouped characters."""
new_lst = []
count = 1
for i in range(len(lst) - 1): # we range to the second to last index since we're checking if lst[i] == lst[i + 1].
if lst[i] == lst[i + 1]:
count += 1
else:
new_lst.append([lst[i],count]) # Create a lst of lists. Each list contains a character and the count of adjacent identical characters.
count = 1
new_lst.append((lst[-1],count)) # Return the last character (we didn't reach it with our for loop since indexing until second to last).
grouped = [char*count for [char, count] in new_lst]
return grouped
def compress_group(string):
"""Returns a compressed two character string containing a character and a number.
Takes in a string of identical characters and returns the compressed string
consisting of the character and the length of the original string.
Example
-------
"AAA"-->"A3"
Parameters:
-----------
Input:
string: str
A string of identical characters.
Output:
------
compressed_str: str
A compressed string of length two containing a character and a number.
"""
return str(string[0]) + str(len(string))
def compress(string):
"""Returns a compressed representation of a string.
Compresses the string by mapping each run of identical characters to a
single character and a count.
Ex.
--
compress('AAABBCDDD')--> 'A3B2C1D3'.
Only compresses string if the compression is shorter than the original string.
Ex.
--
compress('A')--> 'A' # not 'A1'.
Parameters
----------
Input:
string: str
The string to compress
Output:
compressed: str
The compressed representation of the string.
"""
try:
split_str = [char for char in string] # Create list of single characters.
grouped = groupby_char(split_str) # Group characters if characters are identical.
compressed = ''.join( # Compress each element of the grouped list and join to a string.
[compress_group(elem) for elem in grouped])
if len(compressed) < len(string): # Only return compressed if compressed is actually shorter.
return compressed
else:
return string
except IndexError: # If our input string is empty, return an empty string.
return ""
except TypeError: # If we get something that's not compressible (including NoneType) return None.
return None
In [69]:
# %load ../scripts/compress/compress_tests.py
# This will fail to run because in wrong directory
from compress.compressor import *
def compress_test():
assert compress('AAABBCDDD') == 'A3B2C1D3'
assert compress('A') == 'A'
assert compress('') == ''
assert compress('AABBCC') == 'AABBCC' # compressing doesn't shorten string so just return string.
assert compress(None) == None
def groupby_char_test():
assert groupby_char(["A", "A", "A", "B", "B"]) == ["AAA", "BB"]
def compress_group_test():
assert compress_group("AAA") == "A3"
assert compress_group("A") == "A1"