Question

我有这个值来自sql查询到postgresql（我更改了行的颜色以显示firts值是相同的）

**('80255', 'GRN', Decimal('4.00000000000000000000'))**
('80425', 'GRN', Decimal('1.00000000000000000000'))
**('80255', 'BCN', Decimal('1.00000000000000000000'))**
('80425', 'BCN', Decimal('22.0000000000000000'))
**('80255', 'PT', Decimal('2.0000000000000000'))**
('80425', 'PT', Decimal('5.0000000000000000'))
...

我想转换到下一个，转换为csv文件。

80255;4;1;2
80425;1;22;5

对于每个sku（如80425），我想将库存放在每个位置（GRN / BCN / PT）

Answer 1

当您拥有具有多个值的键时使用的一个好工具是collections.defaultdict defaultdict容器为每个新密钥创建一个新对象，当它遇到已存在的密钥时，它就会运行那把钥匙;在这种情况下，操作是列表方法.append()。

然后您可以使用csv模块

写入文件

import csv, collections as co, decimal as dc, itertools as it

fromdb = [('80255', 'GRN', dc.Decimal('4.00000000000000000000'))
         ,('80425', 'GRN', dc.Decimal('1.00000000000000000000'))
         ,('80255', 'BCN', dc.Decimal('1.00000000000000000000'))
         ,('80425', 'BCN', dc.Decimal('22.0000000000000000'))
         ,('80255', 'PT', dc.Decimal('2.0000000000000000'))
         ,('80425', 'PT', dc.Decimal('5.0000000000000000'))]

tocsv = co.defaultdict(list)
for i in fromdb:
    tocsv[i[0]].append(i[-1])

with open('output.txt','wb') as f:
    cw = csv.writer(f,delimiter=';')
    for k,v in tocsv.items():
        cw.writerow(tuple(it.chain((k,),v)))

Answer 2

继续@ bernie的好回答，这里有一个包含更多注释的变体，它不依赖于行顺序，并且可以处理缺失的值：

# import collections and abbreviate it as co
import collections as co
import decimal as dc
import csv

fromdb = [('80255', 'GRN', dc.Decimal('4.00000000000000000000'))
         ,('80425', 'GRN', dc.Decimal('1.00000000000000000000'))
         ,('80255', 'BCN', dc.Decimal('1.00000000000000000000'))
         ,('80425', 'BCN', dc.Decimal('22.0000000000000000'))
         ,('80255', 'PT', dc.Decimal('2.0000000000000000'))
         ,('80425', 'PT', dc.Decimal('5.0000000000000000'))]

# Build a dictionary that maps values like '80255'
# to a dictionary containing values for GRN, BCN and PT
tocsv = co.defaultdict(lambda:
    {'GRN': 0, 'BCN': 0, 'PT': 0} )
for i in fromdb:
    # Array index -1 refers to the last element
    # For example, set tocsv['80255']['GRN'] to 4.00
    tocsv[i[0]][i[1]] = i[-1]

with open('output.txt','wb') as f:
    cw = csv.writer(f,delimiter=';')
    # items() returns an iterable of key (k) and value (v) pairs
    # For key '80255', the value will be {'GRN': 4, 'BCN': 1, 'PT': 2}
    for k,v in tocsv.items():
        # writerow() takes an array argument, hence the [] square brackets
        cw.writerow([k, v['GRN'], v['BCN'], v['PT']])

如何将列表转换为csv？

2 个答案: