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TA贡献2037条经验 获得超6个赞
试试这个,通过从中提取.month_name()和 year( .dt.year)df1并将其与df2
>>> df1
index A B C
0 01/01/2010 15 20 30
1 15/01/2010 12 15 25
2 17/02/2010 14 13 35
3 19/02/2010 11 10 22
>>> df2
index year month price
0 0 2010 january 70
1 1 2010 february 80
# merging df1 and df2 by month and year.
>>> df1.merge(df2,
left_on = [pd.to_datetime(df1['index']).dt.year,
pd.to_datetime(df1['index']).dt.month_name().str.lower()],
right_on = ['year', 'month'])
输出:
index_x A B C index_y year month price
0 01/01/2010 15 20 30 0 2010 january 70
1 15/01/2010 12 15 25 0 2010 january 70
2 17/02/2010 14 13 35 1 2010 february 80
3 19/02/2010 11 10 22 1 2010 february 80
TA贡献1785条经验 获得超4个赞
这是愚蠢的答案!我相信你可以做得比这更聪明:) 但考虑到你的表是一个字典列表(你可以轻松地将你的 SQL 表转换为这种格式),这很有效。我知道这不是一个干净的解决方案,但是您要求一个简单的解决方案,这可能是最容易理解的:)
months = {'january': "01",
'february': "02",
'march': "03",
'april':"04",
'may': "05",
'june': "06",
'july': "07",
'august': "08",
'september': "09",
'october': "10",
'november': "11",
'december': "12"}
table1 = [{'index': '01/01/2010', 'A': 15, 'B': 20, 'C': 30},
{'index': '15/01/2010', 'A': 12, 'B': 15, 'C': 25},
{'index': '17/02/2010', 'A': 14, 'B': 13, 'C': 35},
{'index': '19/02/2010', 'A': 11, 'B': 10, 'C': 22}]
table2 = [{'index': 0, 'year': 2010, 'month': 'january', 'price':70},
{'index': 1, 'year': 2010, 'month': 'february', 'price':80}]
def joiner(table1, table2):
for row in table2:
row['tempDate'] = "{0}/{1}".format(months[row['month']], str(row['year']))
for row in table1:
row['tempDate'] = row['index'][3:]
table3 = []
for row1 in table1:
row3 = row1.copy()
for row2 in table2:
if row2['tempDate'] == row1['tempDate']:
row3['price'] = row2['price']
break
table3.append(row3)
return(table3)
table3 = joiner(table1, table2)
print(table3)
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