python字典中的求和

时间:2017-10-20 12:54:06

标签: python python-2.7 python-3.x

如何使用for循环

在python中创建字典
CPO     1
CL      1
SL      1
EL      1
CPO     1
SL      1
CPO     1

因此预期结果应如下所示 { 'CPO':3 'CL':1, 'SL':2 'EL':1}

我试过了:

avail = defaultdict(list)
    cpo = cl = sl= el = 0
    for i in hr_line_id:
        if i.leave_code == 'CPO':
            cpo = cpo + i.no_of_days
            avail['cpo'].append(cpo)
        elif i.leave_code == 'CL':
            cl = cl + i.no_of_days
            avail['cl'].append(cl)
        elif i.leave_code == 'SL':
            sl = sl + i.no_of_days
            avail['sl'].append(sl)
    print avail

5 个答案:

答案 0 :(得分:5)

如@JeanFrancoisFabre所述,这是collections.Counter的完美示例:

from collections import Counter

text = """CPO     1
CL      1
SL      1
EL      1
CPO     1
SL      1
CPO     1"""

count = Counter()

for line in text.split("\n"):
    k,v = line.split()
    count[k] += int(v)

print(count)
Counter({'CPO': 3, 'SL': 2, 'CL': 1, 'EL': 1})

如果您需要小写密钥,可以使用count[k.lower()] += int(v)

Counter({'cpo': 3, 'sl': 2, 'cl': 1, 'el': 1})

如果数量始终为1,您只需编写一行代码:

Counter(line.split()[0] for line in text.split("\n"))
# Counter({'CPO': 3, 'SL': 2, 'CL': 1, 'EL': 1})

答案 1 :(得分:4)

我在评论中使用collections.Counter,但要调整当前代码,这样的事情应该有效:

avail = defaultdict(lambda: 0)
for i in hr_line_id:
    if i.leave_code == 'CPO':
        avail['cpo'] += i.no_of_days
    elif i.leave_code == 'CL':
        avail['cl'] += i.no_of_days
    elif i.leave_code == 'SL':
        avail['sl'] += i.no_of_days
print avail

根据评论,这里的if链增加了很多噪音。假设目标键不是输入的函数(如.lower())和/或您只想允许某组键,这样的东西可能是首选:

avail = defaultdict(lambda: 0)
keyMapping = {
  'CPO': 'cpo',
  'CL' : 'cl',
  'SL' : 'sl'
}
for i in hr_line_id:
  if i.leave_code in keyMapping:
    avail[keyMapping[i.leave_code]] += i.no_of_days
  else:
    pass # handle unexpected key
print avail

答案 2 :(得分:1)

使用Counter的这个稍短的版本怎么样?

from collections import Counter

counts = Counter((i.leave_code.lower(), i.no_of_days) for i in hr_line_id)

答案 3 :(得分:1)

以下是使用循环执行此操作的另一种方法:

text = """CPO     1
CL      1
SL      1
EL      1
CPO     1
SL      1
CPO     1"""

text_split = text.split() # split the text [CPO, 1, CL, 1 and so on]
dict_k = text_split[0::2] # values at odd index
dict_v = text_split[1::2] # values at even index

dict_t = {}

for k, v in zip(dict_k, dict_v):
    dict_t[k] = dict_t.get(k, 0) + int(v)

print(dict_t)

# output:{'SL': 2, 'CPO': 3, 'EL': 1, 'CL': 1}

答案 4 :(得分:0)

除了计数器,你可以简单地循环。

avail = {}
for i in hr_line_id:
    if i.leave_code in avail:
        avail[i.leave_code] += i.no_of_days
    else
        avail[i.leave_code] = i.no_of_days