Python 元组列表,按第一个元素分组并添加第二个元素
假设我有以下元组列表:Python 元组列表,按第一个元素分组并添加第二个元素,python,list,tuples,Python,List,Tuples,假设我有以下元组列表: tuples = [ [ ('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00') ], [ ('2017-04-12', '47294
tuples = [
[
('2017-04-11', '2000000.00'),
('2017-04-12', '1000000.00'),
('2017-04-13', '3000000.00')
],
[
('2017-04-12', '472943.00'),
('2017-04-13', '1000000.00')
]
# ...
]
我如何根据第一个元素(日期)对它们进行分组并添加另一个元素
例如,我想要这样的东西:
tuples = [('2017-04-11', '2000000.00'), ('2017-04-12', '1472943.00'), ('2017-04-13', '4000000.00')],
使用
itertools.chain.from_iterable
、itertools.groupby
和sum
函数的解决方案:
import itertools, operator
tuples = [
[('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
[('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
]
result = [(k, "%.2f" % sum(float(t[1]) for t in g))
for k,g in itertools.groupby(sorted(itertools.chain.from_iterable(tuples)), operator.itemgetter(0))]
print(result)
输出:
[('2017-04-11', '2000000.00'), ('2017-04-12', '1472943.00'), ('2017-04-13', '4000000.00')]
我的方法是将嵌套列表转换为平面列表并进行迭代:
t = [
[('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
[('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
]
a={}
for i,j in sum(t,[]):
a[i]=a[i]+float(j) if i in a else float(j)
print(a)
输出:
{'2017-04-11': 2000000.0, '2017-04-13': 4000000.0, '2017-04-12': 1472943.0}
如果需要列表,可以使用
[(k,v)表示a.items()中的k,v]。
展平列表,然后使用defaultdict:
from collections import defaultdict
flattened_tuples = [item for sublist in tuples for item in sublist]
result = defaultdict(float)
for date, value in flattened_tuples:
result[date] += float(value)
print(result)
返回
{'2017-04-11':2000000.0,'2017-04-12':1472943.0,'2017-04-13':4000000.0}
首先,从元组列表中展开元组列表,然后使用
因此,您已经发布了您的输入和预期结果。但是你自己解决问题的方法是什么?请发布您的代码,并告诉我们您在哪里被卡住了。看到预期的输出无法理解清楚。但是:将组标识作为键,并用它创建一个字典<代码>{“组标识”,[上面提到的列表]}。使用python以这种格式创建数据。这就是你想要的吗?我问了一个问题,因为我尝试了很多解决方案,但都没能成功。我尝试了map,itertools.izip_,但没有任何运气@mpf82
import itertools
import operator
lists = [
[('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
[('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
]
# Step 1: Flat a list of tuples out of a list of lists of tuples
list_tuples = [t for sublist in lists for t in sublist]
'''
[('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00'), ('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
'''
# Step 2: Groupby
results = list()
for key, group in itertools.groupby(sorted(list_tuples), operator.itemgetter(0)):
s = sum(float(t[1]) for t in group)
results.append((key, s))
print(results)
#[('2017-04-11', 2000000.0), ('2017-04-12', 1472943.0), ('2017-04-13', 4000000.0)]