Pandas计算groupby对象内的连续日期观察值

Question

这是我正在使用的数据框的一个例子:

d = {
'item_number':['bdsm1000', 'bdsm1000', 'bdsm1000', 'ZZRWB18','ZZRWB18', 'ZZRWB18', 'ZZRWB18', 'ZZHP1427BLK', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1427', 'ZZHP1414', 'ZZHP1414', 'ZZHP1414', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115WNTR', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE', 'WRM115SCFRE'],
'Comp_ID':[2454, 2454, 2454, 1395, 1395, 1395, 1395, 3378, 1266941, 660867, 43978, 1266941, 660867, 43978, 1266941, 660867, 43978, 1266941, 660867, 43978, 43978, 43978, 43978, 1197347907, 70745, 4737, 1197347907, 4737, 1197347907, 70745, 4737, 1197347907, 70745, 4737, 1197347907, 4737, 1197487704, 1197347907, 70745, 23872, 4737, 1197347907, 4737, 1197487704, 1197347907, 23872, 4737, 1197487704, 1197347907, 70745],
'date':['2016-11-22', '2016-11-20', '2016-11-19', '2016-11-22', '2016-11-20', '2016-11-19', '2016-11-18', '2016-11-22', '2016-11-22', '2016-11-22', '2016-11-22', '2016-11-20', '2016-11-20', '2016-11-20', '2016-11-19', '2016-11-19', '2016-11-19', '2016-11-18', '2016-11-18', '2016-11-18', '2016-11-22', '2016-11-20', '2016-11-19', '2016-11-22', '2016-11-22', '2016-11-22', '2016-11-21', '2016-11-21', '2016-11-20', '2016-11-20', '2016-11-20', '2016-11-19', '2016-11-19', '2016-11-19', '2016-11-18', '2016-11-18', '2016-11-22', '2016-11-22', '2016-11-22', '2016-11-22', '2016-11-22', '2016-11-21', '2016-11-21', '2016-11-20', '2016-11-20', '2016-11-20', '2016-11-20', '2016-11-19', '2016-11-19', '2016-11-19']}

df = pd.DataFrame(data=d)
df.date = pd.to_datetime(df.date)

我想计算从2016-11-22开始的连续观察,即按Comp_ID和item_number分组.

基本上,我要做的是,计算连续多少天有一个观察值从今天的日期开始计算每个Comp_ID和item_number.(这个例子是在11月22日放在一起的)在今天之前几天/几天观察到的连续观察是不相关的.只有像今天这样的序列......昨天......前天......等等都是相关的.

我让这个工作在一个较小的样本上,但它似乎在更大的数据集上被绊倒.

这是较小样本的代码.我需要找到数千个卖家/商品的连续日期.出于某种原因,下面的代码不适用于较大的数据集.

d = {'item_number':['KIN005','KIN005','KIN005','KIN005','KIN005','A789B','A789B','A789B','G123H','G123H','G123H'],
'Comp_ID':['1395','1395','1395','1395','1395','7787','7787','7787','1395','1395','1395'],
'date':['2016-11-22','2016-11-21','2016-11-20','2016-11-14','2016-11-13','2016-11-22','2016-11-21','2016-11-12','2016-11-22','2016-11-21','2016-11-08']}

df = pd.DataFrame(data=d)
df.date = pd.to_datetime(df.date)
d = pd.Timedelta(1, 'D')

df = df.sort_values(['item_number','date','Comp_ID'],ascending=False)

g = df.groupby(['Comp_ID','item_number'])
sequence = g['date'].apply(lambda x: x.diff().fillna(0).abs().le(d)).reset_index()
sequence.set_index('index',inplace=True)
test = df.join(sequence)
test.columns = ['Comp_ID','date','item_number','consecutive']
g = test.groupby(['Comp_ID','item_number'])
g['consecutive'].apply(lambda x: x.idxmin() - x.idxmax() )

这会为较小的数据集获得所需的结果:

Comp_ID  item_number
1395     G123H          2
         KIN005         3
7787     KIN005         2
Name: consecutive, dtype: int64

Answer 1

您可以这样操作：

today = pd.to_datetime('2016-11-22')

# sort DF by `date` (descending)    
x = df.sort_values('date', ascending=0)
g = x.groupby(['Comp_ID','item_number'])
# compare the # of days to `today` with a consecutive day# in each group
x[(today - x['date']).dt.days == g.cumcount()].groupby(['Comp_ID','item_number']).size()

结果：

Comp_ID  item_number
1395     G123H          2
         KIN005         3
7787     A789B          2
dtype: int64

PS感谢@DataSwede的更快的diff计算！

说明：

In [124]: x[(today - x['date']).dt.days == g.cumcount()] \
           .sort_values(['Comp_ID','item_number','date'], ascending=[1,1,0])
Out[124]:
  Comp_ID       date item_number
8    1395 2016-11-22       G123H
9    1395 2016-11-21       G123H
0    1395 2016-11-22      KIN005
1    1395 2016-11-21      KIN005
2    1395 2016-11-20      KIN005
5    7787 2016-11-22       A789B
6    7787 2016-11-21       A789B