Question

['2016-01-01 00:00:00', '0',
 '2016-01-01 08:00:00', '268705.0',
 '2016-01-01 16:00:00', '0',
 '2016-01-02 00:00:00', '0',
 '2016-01-02 08:00:00', '0.0',
 '2016-01-02 16:00:00', '0.0',
 '2016-01-03 00:00:00' ...
 ... etc for 1 year]

基本上，我将日期和能量产生作为一个整数。我要使它看起来像

['2016-01-01 00:00:00;0',
 '2016-01-01 08:00:00;268705.0',
 ... etc]

又名['date;energy']

有什么提示吗？我对此并不陌生，需要它来完成我的课程...

Answer 1

使用zip()和列表切片可从索引0开始压缩每个第二元素，而从索引1开始压缩每个第二元素。

data = ['2016-01-01 00:00:00', '0',
        '2016-01-01 08:00:00', '268705.0',
        '2016-01-01 16:00:00', '0',
        '2016-01-02 00:00:00', '0',
        '2016-01-02 08:00:00', '0.0',
        '2016-01-02 16:00:00', '0.0',
        '2016-01-03 00:00:00', "18.05",]

new_data = list(zip(data[0::2],data[1::2]))

print(new_data)

combined = ["{};{}".format(a,b) for a,b in new_data]

print(combined)

输出：

# new_data (I would vouch to use that further on)
[('2016-01-01 00:00:00', '0'), ('2016-01-01 08:00:00', '268705.0'), 
 ('2016-01-01 16:00:00', '0'), ('2016-01-02 00:00:00', '0'), 
 ('2016-01-02 08:00:00', '0.0'), ('2016-01-02 16:00:00', '0.0'), 
 ('2016-01-03 00:00:00', '18.05')]

# combined
['2016-01-01 00:00:00;0', '2016-01-01 08:00:00;268705.0', '2016-01-01 16:00:00;0',
 '2016-01-02 00:00:00;0', '2016-01-02 08:00:00;0.0', '2016-01-02 16:00:00;0.0', 
 '2016-01-03 00:00:00;18.05']

如果我是你，我不会str-combine他们，而是进一步使用元组。 F.e.如果您想每天汇总能量：

new_data = sorted((zip(data[0::2],data[1::2])))

from itertools import groupby 

# x[0][0:10] uses the 1st element of each tuple (the datetime) and slices only the date
# from it. That is used to group all data.
k = groupby(new_data, lambda x:x[0][0:10])

summs = []
for date,tups in k:
    summs.append( (date,sum(float(x[1]) for x in tups)) )

print(summs)

输出：

[('2016-01-01', 268705.0), ('2016-01-02', 0.0), ('2016-01-03', 18.05)]

Answer 2

使用zip生成所需值的压缩列表，然后使用join将列表与;结合起来

>>> a=['2016-01-01 00:00:00', '0', '2016-01-01 08:00:00', '268705.0', '2016-01-01 16:00:00', '0']
>>> [';'.join(i) for i in zip(a[::2],a[1::2])]
['2016-01-01 00:00:00;0', '2016-01-01 08:00:00;268705.0', '2016-01-01 16:00:00;0']
>>>

在列表中合并两个字符串

2 个答案: