Python:如何查找重复项并求和它们的值?
例如,我有上面的元组列表,如何找到重复的元组(即“Visa”)并求和它们的值(即980.5+215.0)?输出应为:Python:如何查找重复项并求和它们的值?,python,list,python-3.x,Python,List,Python 3.x,例如,我有上面的元组列表,如何找到重复的元组(即“Visa”)并求和它们的值(即980.5+215.0)?输出应为: [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)] 使用字典: [('Visa', 1195.5), ('Rogers', 61.5)] 使用字典: [('Visa', 1195.5), ('Rogers', 61.5)] 如果你想保持列表的顺序,那么我建议使用字典 data = [('Visa', 980.5), ('
[('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
使用字典:
[('Visa', 1195.5), ('Rogers', 61.5)]
使用字典:
[('Visa', 1195.5), ('Rogers', 61.5)]
如果你想保持列表的顺序,那么我建议使用字典
data = [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
sum = {}
for item in data:
if not item[0] in sum:
sum[ item[0] ] = 0
sum[ item[0] ] += item[1]
print sum.items()
如果你想保持列表的顺序,那么我建议使用字典
data = [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
sum = {}
for item in data:
if not item[0] in sum:
sum[ item[0] ] = 0
sum[ item[0] ] += item[1]
print sum.items()
使用集合:
lst = [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
lst2 = [(tup[0], sum([val for n, val in lst if n == tup[0]])) for tup in lst]
res = []
for tup in lst2:
if tup not in res:
res.append(tup)
print(res)
li=[('Rogers', 10), ('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
s=set([i[0] for i in li])
x=[]
for i in s:
sum=0
for j in li:
if i == j[0]:
sum+=j[1]
x.append(sum)
final_list=zip(s,x)
print final_list
输出:
lst = [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
lst2 = [(tup[0], sum([val for n, val in lst if n == tup[0]])) for tup in lst]
res = []
for tup in lst2:
if tup not in res:
res.append(tup)
print(res)
li=[('Rogers', 10), ('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
s=set([i[0] for i in li])
x=[]
for i in s:
sum=0
for j in li:
if i == j[0]:
sum+=j[1]
x.append(sum)
final_list=zip(s,x)
print final_list
使用集合:
lst = [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
lst2 = [(tup[0], sum([val for n, val in lst if n == tup[0]])) for tup in lst]
res = []
for tup in lst2:
if tup not in res:
res.append(tup)
print(res)
li=[('Rogers', 10), ('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
s=set([i[0] for i in li])
x=[]
for i in s:
sum=0
for j in li:
if i == j[0]:
sum+=j[1]
x.append(sum)
final_list=zip(s,x)
print final_list
输出:
lst = [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
lst2 = [(tup[0], sum([val for n, val in lst if n == tup[0]])) for tup in lst]
res = []
for tup in lst2:
if tup not in res:
res.append(tup)
print(res)
li=[('Rogers', 10), ('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
s=set([i[0] for i in li])
x=[]
for i in s:
sum=0
for j in li:
if i == j[0]:
sum+=j[1]
x.append(sum)
final_list=zip(s,x)
print final_list
看来大家都忘了:
看来大家都忘了:
A
collections.defaultdict
将是最有效的方法:
from collections import Counter
c = Counter()
for card, val in lst:
c.update({card: val})
print(list(c.items()))
# [('Visa', 1195.5), ('Rogers', 61.5)]
输出:
from collections import defaultdict
l= [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
d = defaultdict(float)
for k,v in l:
d[k] += v
defaultdict(,{'Visa':1195.5,'Rogers':61.5})
Acollections.defaultdict
将是最有效的方法:
from collections import Counter
c = Counter()
for card, val in lst:
c.update({card: val})
print(list(c.items()))
# [('Visa', 1195.5), ('Rogers', 61.5)]
输出:
from collections import defaultdict
l= [('Visa', 980.5), ('Rogers', 61.5), ('Visa', 215.0)]
d = defaultdict(float)
for k,v in l:
d[k] += v
defaultdict(,{'Visa':1195.5,'Rogers':61.5})
要添加到这个答案中,如果不需要打印,那么列表(sum.items())
就可以了。(sum.items()。(sum.items()。