[英]How to sum the second elements of the list if the first elements in the list are matching
[英]how to sum list elements
list_pairs = str(zip(GetEmpID[row],Duration[row]))
從上面的函數,我得到如下結果。 這里1046,8008,8011
是EmpID
,而1.0,2.3
等是值。
[(u'1046', 1.0)]
[(u'8008', 2.3)]
[(u'8008', 2.2)]
[(u'8011', 1.3)]
我的結果應該如下所示。如果EmpID相同,則添加(sum)該元素。如何在Python中做到這一點。
[(u'1046', 1.0)]
total = 1.0
[(u'8008', 2.3)]
[(u'8008', 2.2)]
total = 4.5
[(u'8011', 1.3)]
total = 1.3
answer = []
for empId, entries in itertools.groupby(sorted(list_pairs, key=operator.itemgetter(0)), key=operator.itemgetter(0)):
answer.append((empId, sum(entry[1] for entry in entries)))
In [17]: list_pairs = [(u'1046', 1.0), (u'8008', 2.3), (u'8008', 2.2), (u'8011', 1.3)]
In [18]: answer = []
In [19]: for empId, entries in itertools.groupby(sorted(list_pairs, key=operator.itemgetter(0)), key=operator.itemgetter(0)):
....: answer.append((empId, sum(entry[1] for entry in entries)))
....:
In [20]: answer
Out[20]: [(u'1046', 1.0), (u'8008', 4.5), (u'8011', 1.3)]
為了使其可讀 :
answer = []
list_pairs.sort(key=operator.itemgetter(0))
groups = itertools.groupby(list_pairs, key=operator.itemgetter(0))
for empId, entries in groups:
answer.append((empId, sum(entry[1] for entry in entries)))
保留一個具有當前EmpID的變量以及該EmpID的最新值的總和。 當EmpID更改時,將其輸出(或將其保存到列表中,然后插入)
list_pairs = str(zip(GetEmpID[row],Duration[row]))
last=""
last_sum=0
for empid, value in list_pairs:
if empid!=last:
if last:
print last, last_sum
last, last_sum= empid, 0
else:
last_sum+=value
print last, last_sum
如果需要通過單行執行輸出,請嘗試此操作。
data=[(u'1046', 1.0), (u'8008', 2.2999999999999998), (u'8008', 2.2000000000000002), (u'8011', 1.3)]
import itertools
[(key, sum(x for _,x in value))for key, value in itertools.groupby(data, lambda x: x[0])]
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