如何获取所有相同的"key[0]"并在字典中计算其值

Question

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 函数可以成功运行，但由于 orderrow 函数适用于集群 All 金额，从"A"开始，然后对行进行排名("A"、"B"、"C"、"D"(

Answer 1

您可以使用sorted并直接在counts上应用

import operator
sorted_x = sorted(counts.items(), key=operator.itemgetter(1), reverse=True)

您可以采用新dict并分配键，值如下所示：

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}

替换函数将如下所示，

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