如何在python中计算周差并使用count添加缺少的周数
我有一个这样的数据框,我必须得到缺失的周值,并在它们之间计数如何在python中计算周差并使用count添加缺少的周数,python,pandas,Python,Pandas,我有一个这样的数据框,我必须得到缺失的周值,并在它们之间计数 year Data Id 20180406 57170 A 20180413 55150 A 20180420 51109 A 20180427 57170 A 20180504 55150 A 20180525 51109 A 输出应该是这样的 Id Start year end-year count A 20180420 20180420
year Data Id
20180406 57170 A
20180413 55150 A
20180420 51109 A
20180427 57170 A
20180504 55150 A
20180525 51109 A
输出应该是这样的
Id Start year end-year count
A 20180420 20180420 1
A 20180518 20180525 2
使用:
我的输出有点不同,你能检查一下吗?你的答案是正确的。我们可以获得季度数据吗?对于查询年份数据Id 2019Q4 57170 A 2019Q3 55150 A 2019Q2 51109 A 2019Q1 51109 A 2018Q1 57170 B 2018Q4 55150 B 2017Q4 51109 C 2017Q2 51109 C 2017Q1 51109 C Id,我在同一问题中还有另一个问题开始年份结束年份计数B 2018Q2 2018Q3 2 B 2017Q3 2018Q3 1我如何使用python Pandapl实现这一点请提出新问题:)我已经发布了问题,请帮助我。。如何降低时间复杂度或提高程序的效率使用python查找月间隔
#converting to week period starts in Thursday
df['year'] = pd.to_datetime(df['year'], format='%Y%m%d').dt.to_period('W-Thu')
#resample by start of months with asfreq
df1 = (df.set_index('year')
.groupby('Id')['Id']
.resample('W-Thu')
.asfreq()
.rename('val')
.reset_index())
print (df1)
Id year val
0 A 2018-04-06/2018-04-12 A
1 A 2018-04-13/2018-04-19 A
2 A 2018-04-20/2018-04-26 A
3 A 2018-04-27/2018-05-03 A
4 A 2018-05-04/2018-05-10 A
5 A 2018-05-11/2018-05-17 NaN
6 A 2018-05-18/2018-05-24 NaN
7 A 2018-05-25/2018-05-31 A
#onverting to datetimes with starts dates
#http://pandas.pydata.org/pandas-docs/stable/timeseries.html#converting-between-representations
df1['year'] = df1['year'].dt.to_timestamp('D', how='s')
print (df1)
Id year val
0 A 2018-04-06 A
1 A 2018-04-13 A
2 A 2018-04-20 A
3 A 2018-04-27 A
4 A 2018-05-04 A
5 A 2018-05-11 NaN
6 A 2018-05-18 NaN
7 A 2018-05-25 A
m = df1['val'].notnull().rename('g')
#create index by cumulative sum for unique groups for consecutive NaNs
df1.index = m.cumsum()
#filter only NaNs row and aggregate first, last and count.
df2 = (df1[~m.values].groupby(['Id', 'g'])['year']
.agg(['first','last','size'])
.reset_index(level=1, drop=True)
.reset_index())
print (df2)
Id first last size
0 A 2018-05-11 2018-05-18 2