added solutions

Author: panzerballett

README.md

@@ -1,7 +1,15 @@
 # pydebug
-a small set of python sketches that do not run as expected and wait to be debugged.
+a small set of python sketches that do not run as expected and wait to be
+debugged.
 
 used as exercises for
+
 ![power.coders](powercoders-cropped.png "power.coders")
 
-good luck!
+the idea of these buggy sketches is to learn to read python's traceback. python
+will tell you very precisely why and also where it is unhappy with your code.
+one of the goals of these snippets is to learn to read the traceback and then
+fix the code.
+
+
+good luck!

src/01.py problems/01.py

@@ -2,11 +2,15 @@
 
 '''
 this code is buggy. make it run!
+
+
+it is supposed to print number from 0 up to (and without) 10 and then print
+if that number is odd or even.
 '''
 
 for i in range(10):
      print(i)
-    if i%2 == 0
+    if i % 2 == 0
         print('even!')
     else
         print('odd')

src/02.py problems/02.py

@@ -2,6 +2,10 @@
 
 '''
 again, buggy code. can you guess what it should do and fix?
+
+
+first we want to print out the value of the key 'b'; then we want to print
+out all the values (1, 2, 3).
 '''
 
 dct = {'a': 1, 'b': 2, 'c': 3}

src/03.py problems/03.py

@@ -1,7 +1,10 @@
 #!/usr/bin/env python3
 
 '''
-bugs.
+bugs again.
+
+this should print the user name given in the command line. but it does not.
+please fix!
 '''
 
 user = input('user name please: ')

src/04.py problems/04.py

@@ -2,6 +2,10 @@
 
 '''
 again, please fix!
+
+first we want to print out the last element of lst.
+
+then we want to print the last 5 elements.
 '''
 
 lst = list(range(10))

problems/05.py

@@ -0,0 +1,18 @@
+#!/usr/bin/env python3
+
+'''
+do *not* run this! instead, try to find out what it does without running.
+
+*then* run and see if you were correct.
+
+there is not really a fix for this; just remove or comment out the offending
+line.
+'''
+
+
+def f():
+    a = 5
+    print(a)
+
+f()
+print(a)

problems/06.py

@@ -0,0 +1,19 @@
+#!/usr/bin/env python3
+
+'''
+we want to remove all integers other than 5 from a list.
+
+run the code, be amazed and try to come up with a solution.
+
+don't hurt yourself too much if you can't make this one work. go have a look
+at the solution once you have exhausted all your ideas.
+'''
+
+lst = list(range(10))
+
+for n in lst:
+    if n != 5:
+        lst.remove(n)
+
+# this should print a list containing only the number 5
+print(lst)

src/07.py problems/07.py

@@ -2,6 +2,8 @@
 
 '''
 why is nothing happening? why is there no error?
+
+we want to call function...
 '''
 
 

problems/08.py

@@ -0,0 +1,9 @@
+#!/usr/bin/env python3
+
+'''
+just fix please. take a *close* look at the traceback. you will see this a lot!
+'''
+
+variable = 'variable'
+
+print(varaible)

src/09.py problems/09.py

@@ -1,7 +1,7 @@
 #!/usr/bin/env python3
 
 '''
-again, look at the traceback and fix!
+again, look closely at the traceback and fix!
 '''
 
 

src/10.py problems/10.py

@@ -4,7 +4,7 @@
 trying to create a datetime. see the documentation:
 https://docs.python.org/3/library/datetime.html
 
-why does this not work?!
+why does this not work?! please fix.
 '''
 
 import datetime

problems/11.py

@@ -0,0 +1,27 @@
+#!/usr/bin/env python3
+
+'''
+this will show you a python feature that is important to be aware of. we
+pretend to be ignorant of the feautre to show how it works. can you fix it for
+us?
+
+the function f is supposed to get one argument and return the argument as
+only item in a list. (of course we could just have returned '[x]' or 'list(x)'
+but we want to use append for this one).
+
+look at what is happening... can you fix that without changing the last 2 lines
+of f?
+'''
+
+
+def f(x, lst=[]):
+    '''
+    append x to lst.
+    '''
+
+    lst.append(x)  # do not modify this line
+    return lst     # do not modify this line
+
+print(f(0))    # expecting [0]
+print(f('a'))  # expecting ['a']
+print(f([]))   # expecting [[]]

problems/12.py

@@ -0,0 +1,45 @@
+#!/usr/bin/env python3
+
+'''
+now we are going to use the python feature we just learned about in the
+previous example to cache values of a function.
+
+suppose long_running_f is a function that runs for quite some time (we simulate
+that here with time.sleep). assume this function is called many times but the
+number of different arguments is small.
+
+we do not want to repeat the long calculation (or data retrival/...). so we
+store {argument: return_value} in a dictionary and just return the value if we
+have calculated it already. otherwise we calculate the value, add it to the
+cache and return it.
+
+the code does not work as expected. please fix it!
+'''
+
+
+import time
+
+
+def long_running_f(x, _cache={}):
+    """
+    computes the power of 2 for a number x.
+
+    Keeps results in memory because computing powers is expensive.
+    """
+    # if the result for x is not yet in memory, we compute it
+    if _cache[x] is None:
+        print('- calculating value for {}'.format(x))
+        time.sleep(2)  # we preted this takes some time
+        ret = 2 ** x
+        _cache[x] = ret
+    else:
+        print('- can get value directly from cache for {}'.format(x))
+        ret = _cache[x]
+
+    return ret
+
+
+for arg in (0, 1, -5, 3, 0, -3, 5, 3, 1, 2, 1, 4, 1, 4, -5, -3, 1, 5):
+
+    ret = long_running_f(arg)
+    print('long_running_f({:2d}) = {}'.format(arg, ret))

solutions/01.py

@@ -0,0 +1,20 @@
+#!/usr/bin/env python3
+
+'''
+the solution.
+
+there was 1 space too many on the first print line.
+you also could have added a space before 'if' and 'else': this would have
+resulted in correct python code as well. but it is considered bad style to have
+indentations that are not multiples of 4 spaces.
+see: https://www.python.org/dev/peps/pep-0008/
+
+then there were the ':' missing for the 'if/else' statement.
+'''
+
+for i in range(10):
+    print(i)        # indentation was off here: 1 space too many
+    if i % 2 == 0:  # ':' was missing here
+        print('even!')
+    else:           # ':' was missing here
+        print('odd')

solutions/02.py

@@ -0,0 +1,45 @@
+#!/usr/bin/env python3
+
+'''
+dictionaries consist of key/value pairs: {key: value, ...}.
+
+value addess is done this way: dct[key] (and not dct[0] for e.g. the first
+element; dictionaries are unordered! the order in which you define the
+dictionary or the order you see when you are printing it has *no* meaning!)
+
+there are 3 ways to iterate over dictionaries:
+
+    for key in dct:  # iterate over keys
+        print(key)
+
+this is identical to:
+
+    for key in dct.keys():  # iterate over keys
+        print(key)
+
+
+then you can iterate over the values:
+
+    for value in dct.values():  # iterate over values
+        print(value)
+
+and you can directly iterate over key/value pairs:
+
+    for key, value in dct.items():  # iterate over key/value pairs
+        print(key, value)
+
+details here:
+https://docs.python.org/3/tutorial/datastructures.html#dictionaries
+'''
+
+dct = {'a': 1, 'b': 2, 'c': 3}
+
+print("the value of key 'b' is:", dct['b'])
+
+# print out the numbers:
+for value in dct.values():
+    print(value)
+
+# if you need them sorted, you have to state that:
+for value in sorted(dct.values()):
+    print(value)

solutions/03.py

@@ -0,0 +1,9 @@
+#!/usr/bin/env python3
+
+'''
+'user' is a str(ing) and not the name of a variable.
+'''
+
+user = input('user name please: ')
+
+print('got the username:', user)  # removed the quotes.

solutions/04.py

@@ -0,0 +1,31 @@
+#!/usr/bin/env python3
+
+'''
+lst = list(range(10)) means:
+lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
+
+list elements are accessed by their index in the list.
+
+    lst['10']
+
+tries to access a list with a string. that will not work! and the index of the
+last element is not 10 but 9 (indices start at 0).
+
+in general you should access the last element of a list with
+
+    lst[-1]
+
+negative indices count from the end of the list.
+
+
+search for 'slicing' here (or find an explanation for 'slice' in python).
+https://docs.python.org/3/tutorial/introduction.html
+'''
+
+lst = list(range(10))
+
+print('the last element is', lst[-1])
+
+# loop through last 5 elements only (should print numbers 5, 6, 7, 8, 9)
+for i in lst[5:]:  # ':'  was missing: we need a slice and not a single index.
+    print(i)

solutions/05.py

@@ -0,0 +1,14 @@
+#!/usr/bin/env python3
+
+'''
+'a' onlylives in the local namespace of 'f()' and is undefined in the global
+namespace.
+'''
+
+
+def f():
+    a = 5
+    print(a)
+
+f()
+# print(a)  # just comment this out. 'a' is undefined here.

solutions/06.py

@@ -0,0 +1,16 @@
+#!/usr/bin/env python3
+
+'''
+take-home-message: *never* change a list (or anyghint you iterate over) while
+you are looping over it (unless you *really* know what you are doing). this
+will almost always end in tears!
+
+filtering lists is most conveniently done using a list comprehension:
+https://docs.python.org/3/howto/functional.html#generator-expressions-and-list-comprehensions
+
+'''
+
+lst = list(range(10))
+lst = [value for value in lst if value == 5]
+
+print(lst)

solutions/07.py

@@ -0,0 +1,15 @@
+#!/usr/bin/env python3
+
+'''
+the function is never called. 'function' is just a reference to the function
+object of the function. that is valid python - the statement just does not
+do anything.
+
+in order to call the function, you must add '()': function().
+'''
+
+
+def function():
+    print('function called!')
+
+function()  # added '()' to *call* the function.

solutions/08.py

@@ -0,0 +1,10 @@
+#!/usr/bin/env python3
+
+'''
+typo. misspelled 'variable'. did you notice how precisely python tells you
+what went wrong and where? have a look at the traceback again...
+'''
+
+variable = 'variable'
+
+print(variable)