Myujth

I have a 2 Dimensional list and have to get 2 columns from the 2D list and place the values from each column as key:value pairs.

Example:

table = [[15, 29, 6, 2],
 [16, 9, 8, 0],
 [7, 27, 16, 0]]

def averages(table, col, by):

 columns = tuple(([table[i][col] for i in range(len(table))])) #Place col column into tuple so it can be placed into dictionary
 groupby = tuple(([table[i][by] for i in range(len(table))])) #Place groupby column into tuple so it can be placed into dictionary

 avgdict = 
 avgdict[groupby] = [columns]
 print(avgdict)

averages(table, 1, 3)

Output is:

(2, 0, 0): [(29, 9, 27)]

I am trying to get the output to equal:

0:36, 2:29

So essentially the 2 keys of 0 have their values added

I'm having a hard time understanding how to separate each key with their values
and then adding the values together if the keys are equal.

Edit: I'm only using Python Standard library, and not implementing numpy for this problem.

You can create an empty dictionary, then iterate through every element of groupby. If the element in groupby exist in the dictionary, then add the corresponding element in columns to the values in the dictionary. Otherwise, add the element in groupby as key and the corresponding element in columns as value.The implementation is as follows:

table = [[15, 29, 6, 2],
 [16, 9, 8, 0],
 [7, 27, 16, 0]]

def averages(table, col, by):
 columns = tuple(([table[i][col] for i in range(len(table))])) #Place col column into tuple so it can be placed into dictionary
 groupby = tuple(([table[i][by] for i in range(len(table))])) #Place groupby column into tuple so it can be placed into dictionary

 avgdict = 

 for x in range(len(groupby)):
 key = groupby[x]
 if key in avgdict:
 avgdict[key] += columns[x]
 else:
 avgdict[key] = columns[x]

 print(avgdict)

averages(table, 1, 3)

Otherwise, if you want to keep your initial avgdict, then you can change the averages() function to

def averages(table, col, by):
 columns = tuple(([table[i][col] for i in range(len(table))])) #Place col column into tuple so it can be placed into dictionary
 groupby = tuple(([table[i][by] for i in range(len(table))])) #Place groupby column into tuple so it can be placed into dictionary

 avgdict = 
 avgdict[groupby] = [columns]

 newdict = 

 for key in avgdict:
 for x in range(len(key)):
 if key[x] in newdict:
 newdict[key[x]] += avgdict[key][0][x]
 else:
 newdict[key[x]] = avgdict[key][0][x]

 print(newdict)

It took me a minute to figure out what you were trying to accomplish because your function and variable names reference averages but your output is a sum.

Based on your output, it seems you're trying to aggregate row values in a given column by a group in another column.

Here's a recommended solution (which likely could be reduced to a one-liner via list comprehension). This loops through the unique (using set) values (b) in your group by, creates a dictionary key (agg_dict[b]) for the group by being processed, and sums all rows in a given column (col) if the group by is being processed (table[i][by] == by).

table = [[15, 29, 6, 2],
 [16, 9, 8, 0],
 [7, 27, 16, 0]]


def aggregate(tbl, col, by):
 agg_dict = 
 for b in list(set([table[i][by] for i in range(len(table))]))
 agg_dict[b] = sum([table[i][col] for i in range(len(table)) if table[i][by] == b])
 print(agg_dict)


aggregate(table, 1, 3)

Sorry for joining the party late. You can also try the following answer. It doesn't use numpy, and is based on the use of sets to find unique elements in groupby.

table = [[15, 29, 6, 2],
 [16, 9, 8, 0],
 [7, 27, 16, 0]]

def averages(table, col, by):
 columns = tuple(([table[i][col] for i in range(len(table))])) #Place col column into tuple so it can be placed into dictionary
 groupby = tuple(([table[i][by] for i in range(len(table))])) #Place groupby column into tuple so it can be placed into dictionary

 '''groupby_unq: tuple data type 
 stores list of unique entries in groupby.'''
 groupby_unq = tuple(set(groupby))

 '''avg: numpy.ndarray data type
 numpy array of zeros of same length as groupby_unq.''' 
 avg = np.zeros( len(groupby_unq) )

 for i in range(len(groupby)):
 for j in range(len(groupby_unq)):
 if(groupby[i]==groupby_unq[j]): avg[j]+=columns[i]

 avgdict = dict( (groupby_unq[i], avg[i]) for i in range(len(avg)) )

 return avgdict 

result = averages(table, 1, 3)
print result 

0: 36.0, 2: 29.0