counts = {'A1':3,'A2':2,'A3':1,'A4':10,'B1':11,'B2':12,'B3':3,'B4':0,'C1':3,'C2':16,'C3':8,'C4':2}
#Order grid
def orderGrid(grid):
lst = list()
for key,val in grid.items():
lst.append((val,key))
lst.sort(reverse=True)
for val,key in lst:
print key, val
#Order row
def orderRow(row):
count = dict()
for key in row.items():
if key[0] not in count:
count[key] = row[key]
else:
count[key] += row[key]
print 'A:', count
orderGrid function can run successful, but as the orderrow function is for cluster All amount, which start from 'A', and then rank the rows ('A','B','C','D')
You can use sorted
and apply directly on counts
import operator
sorted_x = sorted(counts.items(), key=operator.itemgetter(1), reverse=True)
You can take a new dict
and assign key, values as follows:
In [71]: mydict = {}
In [72]: for k, v in counts.items():
...: if k[0] not in mydict:
...: mydict[k[0]] = v
...: else:
...: mydict[k[0]] += v
...:
In [73]: mydict
Out[73]: {'A': 16, 'B': 26, 'C': 29}
Replacing the functions would look like this,
import operator
counts = {'A1':3,'A2':2,'A3':1,'A4':10,'B1':11,'B2':12,'B3':3,'B4':0,'C1':3,'C2':16,'C3':8,'C4':2}
#Order grid
def orderGrid(grid):
sorted_x = sorted(counts.items(), key=operator.itemgetter(1), reverse=True)
for key,val in sorted_x:
print key, val
#Order row
def orderRow(row):
mydict = {}
for k, v in row.items():
if k[0] not in mydict:
mydict[k[0]] = v
else:
mydict[k[0]] += v
print mydict
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