MySQL：获取计数和平均值[重复](MySQL: Get Counts and Averages [duplicate])

 我认为你需要首先使用boolean indexing进行过滤，然后进行groupby和聚合size 。 
 
 汇总输出并添加reindex以添加由0填充的缺失行： 
 
print (df)
         Date ID
0  01/01/2016  a
1  05/01/2016  a
2  10/05/2017  a
3  05/05/2018  b
4  07/09/2014  b
5  07/09/2014  c
6  12/08/2018  b
 
 
#convert to datetime (if first number is day, add parameter dayfirst)
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
now = pd.datetime.today()
print (now)

oneyarbeforenow =  now - pd.offsets.DateOffset(years=1)
oneyarafternow =  now + pd.offsets.DateOffset(years=1)

#first filter
a = df[df['Date'].between(oneyarbeforenow, now)].groupby('ID').size()
b = df[df['Date'].between(now, oneyarafternow)].groupby('ID').size()
print (a)
ID
a    1
dtype: int64

print (b)
ID
b    2
dtype: int64

df1 = pd.concat([a,b],axis=1).fillna(0).astype(int).reindex(df['ID'].unique(),fill_value=0)
print (df1)
   0  1
a  1  0
b  0  2
c  0  0
 
 编辑： 
 
 如果需要比较每个日期的第一个日期加上或减去每组的year offset需要自定义函数的条件和sum ： 
 
offs = pd.offsets.DateOffset(years=1)

f = lambda x: pd.Series([(x > x.iat[-1] - offs).sum(), \
                        (x < x.iat[-1] + offs).sum()], index=['last','next'])
df = df.groupby('ID')['Date'].apply(f).unstack(fill_value=0).reset_index()
print (df)
  ID  last  next
0  a     1     3
1  b     3     2
2  c     1     1

I think you need between with boolean indexing for filter first and then groupby and aggregate size.
 
Outputs are concated and add reindex for add missing rows filled by 0:
 
print (df)
         Date ID
0  01/01/2016  a
1  05/01/2016  a
2  10/05/2017  a
3  05/05/2018  b
4  07/09/2014  b
5  07/09/2014  c
6  12/08/2018  b
 
 
#convert to datetime (if first number is day, add parameter dayfirst)
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
now = pd.datetime.today()
print (now)

oneyarbeforenow =  now - pd.offsets.DateOffset(years=1)
oneyarafternow =  now + pd.offsets.DateOffset(years=1)

#first filter
a = df[df['Date'].between(oneyarbeforenow, now)].groupby('ID').size()
b = df[df['Date'].between(now, oneyarafternow)].groupby('ID').size()
print (a)
ID
a    1
dtype: int64

print (b)
ID
b    2
dtype: int64

df1 = pd.concat([a,b],axis=1).fillna(0).astype(int).reindex(df['ID'].unique(),fill_value=0)
print (df1)
   0  1
a  1  0
b  0  2
c  0  0
 
EDIT:
 
If need compare each date by first date add or subtract year offset per group need custom function with condition and sum Trues:
 
offs = pd.offsets.DateOffset(years=1)

f = lambda x: pd.Series([(x > x.iat[-1] - offs).sum(), \
                        (x < x.iat[-1] + offs).sum()], index=['last','next'])
df = df.groupby('ID')['Date'].apply(f).unstack(fill_value=0).reset_index()
print (df)
  ID  last  next
0  a     1     3
1  b     3     2
2  c     1     1