1

我有一个如下的数据框：

df = pd.DataFrame({'price': ['480,000,000','477,000,000', '608,700,000', '580,000,000', '350,000,000'], 'sale_date': ['1396/10/30','1396/10/30', '1396/11/01', '1396/11/03', '1396/11/07']})

df
Out[7]: 
         price   sale_date
0  480,000,000  1396/10/30
1  477,000,000  1396/10/30
2  608,700,000  1396/11/01
3  580,000,000  1396/11/04
4  350,000,000  1396/11/04

所以我定义了 period datetime 并按天重新采样

df['sale_date']=df['sale_date'].str.replace('/','').astype(int)
df['price'] = df['price'].str.replace(',','').astype(int)

def conv(x):
    return pd.Period(year=x // 10000,
                     month=x // 100 % 100,
                     day=x % 100, freq='D')
 
df['sale_date'] = df['sale_date'].apply(conv)

s = df.groupby('sale_date')['price'].sum()

那么我想通过前一天的值来填补日期时间的空白。

这是我想要的输出：

In [13]:
         price   sale_date
0  957,000,000  1396/10/30
2  608,700,000  1396/11/01
0  680,000,000  1396/10/02
0  680,000,000  1396/10/03
3  930,000,000  1396/11/04

或通过前一天和第二天
所需的输出：

In [13]: 
         price   sale_date
0  957,000,000  1396/10/30
2  608,700,000  1396/11/01
0  769,000,000  1396/10/02
0  769,000,000  1396/10/03
3  930,000,000  1396/11/04
4

1 回答 1

2

您可以先重新索引而不用0byfill_value参数替换缺失值，然后用 sum byadd和 last 除以转发和填充缺失值2：

df['sale_date']=df['sale_date'].str.replace('/','').astype(int)
df['price'] = df['price'].str.replace(',','').astype(int)

def conv(x):
    return pd.Period(year=x // 10000,
                     month=x // 100 % 100,
                     day=x % 100, freq='D')
 
df['sale_date'] = df['sale_date'].apply(conv)

s = df.groupby('sale_date')['price'].sum()

rng = pd.period_range(s.index.min(), s.index.max(), name='sale_date')
s = s.reindex(rng)
print (s)
sale_date
1396-10-30    957000000.0
1396-10-31            NaN
1396-11-01    608700000.0
1396-11-02            NaN
1396-11-03    580000000.0
1396-11-04            NaN
1396-11-05            NaN
1396-11-06            NaN
1396-11-07    350000000.0
Freq: D, Name: price, dtype: float64

s = s.ffill().add(s.bfill()).div(2).reset_index()
print (s)
    sale_date        price
0  1396-10-30  957000000.0
1  1396-10-31  782850000.0
2  1396-11-01  608700000.0
3  1396-11-02  594350000.0
4  1396-11-03  580000000.0
5  1396-11-04  465000000.0
6  1396-11-05  465000000.0
7  1396-11-06  465000000.0
8  1396-11-07  350000000.0

print ((957000000 + 608700000)/ 2)
782850000.0
于 2020-12-11T08:02:08.990 回答