我的数据框看起来像这样:(不一定是这些日期,长度或顺序)
date1 date2 dummy
2015-10-01 2015-09-02 1
2015-10-01 2015-09-02 1
2015-10-03 2015-09-02 0
2015-10-04 2015-09-05 0
.......... .......... .
.......... .......... .
.......... .......... .
2015-10-20 2015-11-04 1
2015-10-20 2015-11-05 1
我正在创建一个新数据框,其中包含'date2'中最早的日期和'date1'中的最新日期,并填写日期之间的时间段。
startdate = df['date2'].min(axis=0)
enddate = df['date1'].max(axis=0)
def perdelta(start, end, delta):
curr = start
while curr <= end:
yield curr
curr += delta
data2 =[]
for result in perdelta(startdate, enddate, timedelta(days=1)):
data2.append(result)
我想在新数据框中找到每一行日期,将其与'date1'匹配,并计算在'dummy'中有多少个相同的日期。 我可以找到所有的零并用pandas groupby计算它们的特定日期
g = df.groupby(['date1'])
df3 = pd.DataFrame(g.apply(lambda x: x[x['dummy'] == 0]['dummy'].count()), columns=['all_zeros'])
但这只会在'date1'中找到日期并计算零,而不是从我的startdate开始,它也会跳过有一个日期并且不粘贴零的日期(计算非零应该粘贴0)。
我想得到的输出是:
date_newdf count
'startdate' 0 (cuz it does not exist in date1)
2015-09-05 0 (cuz it does not exist in date1)
.......... .
.......... .
.......... .
2015-10-01 3 (found 3 zeroz with the this date)
.......... .
'enddate' 2
等。
复制:
data = {'date1': ['15-10-01', '15-10-01', '15-10-03', '15-10-04', '15-10-05', '15-10-05'],
'date2': ['15-09-02', '15-09-02', '15-09-02', '15-09-05', '15-09-05', '15-09-05'],
'dummy': [1,1,0,0,0,1]}
df = pd.DataFrame(data, columns=['date1', 'date2' , 'dummy'])
答案 0 :(得分:1)
我认为,您需要将reindex函数与列表data2
添加到脚本的末尾,然后将缺少的数据NaN
填充到1
。
输入更好的测试:
date1 date2 dummy
0 2015-10-01 2015-09-02 1
1 2015-10-01 2015-09-02 1
2 2015-10-03 2015-09-02 0
3 2015-10-04 2015-09-05 0
4 2015-10-05 2015-11-05 0
5 2015-10-05 2015-11-05 0
6 2015-10-05 2015-11-05 0
7 2015-10-05 2015-11-05 1
8 2015-10-05 2015-11-05 1
print df3
all_zeros
date1
2015-10-01 0
2015-10-03 1
2015-10-04 1
2015-10-05 3
df3 = df3.reindex(pd.DatetimeIndex(data2))
df3 = df3.fillna(0)
print df3
all_zeros
2015-09-02 0
2015-09-03 0
2015-09-04 0
2015-09-05 0
2015-09-06 0
2015-09-07 0
2015-09-08 0
2015-09-09 0
2015-09-10 0
2015-09-11 0
2015-09-12 0
2015-09-13 0
2015-09-14 0
2015-09-15 0
2015-09-16 0
2015-09-17 0
2015-09-18 0
2015-09-19 0
2015-09-20 0
2015-09-21 0
2015-09-22 0
2015-09-23 0
2015-09-24 0
2015-09-25 0
2015-09-26 0
2015-09-27 0
2015-09-28 0
2015-09-29 0
2015-09-30 0
2015-10-01 0
2015-10-02 0
2015-10-03 1
2015-10-04 1
2015-10-05 3