2 3 # Consider a list of values, xs. Find and return the value at location index

Question

2 3 # Consider a list of values, xs. Find and return the value at location index. 4 # If index is invalid for xs, return response instead. Remember that .get() 5 # for dictionaries works gracefully for non-existing keys. Here we are 6 # implementing a get() function for list type. 7 # -Parameters: 8 # -xs :: list of values of any length. 9 # -index :: an integer, specifying which value in the list to return. 10 # -response :: a python value, to be returned when index is invalid for the 11 # list. Defaults to None. 12 # -Return value: a value from xs at index or the pre-set response 13 # -suggestion : use try-except blocks in your solution. 14 # -Examples: 15# -get(['a','b','c'],0) 16 # -get(('a','b','c'],3) 17 # -get(['a",'b','c'],4,"oops" ) --> 'oops ' 18 19 def get(xs, index, response None): 20 return "not yet implemented" 21 None

Explanation / Answer

def get( xs, index, response=None):
   if( index >= len(xs) or index < 0):
       return response;
   return xs[index];

print get( ['a','b','c'], 0 );
print get( ['a','b','c'], 3 );
print get( ['a','b','c'], 4, "oops" );

def classify( input_string ):
   input_string= input_string.split(" ");
   numbers= [];
   words = [];
   for x in input_string:
       try:
           x = int(x);
           numbers.append(x);
       except:
              if x <> '':
               words.append(x);
   return [ numbers, words ];

def shelve( inventory, product_list ):
   for p in product_list:
       try:
           inventory[ p[0] ] = inventory[ p[0] ] + p[1];
       except:
           inventory[ p[0] ] = p[1];
       if inventory[ p[0] ] < 0:
           raise ValueError('negative amount for '+ p[0] );
   return None;

d = {"kiwi":999};
shelve(d,[("kiwi",-2000)]);

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