使用Python 2.7打开文件a将其转换为字典,其中第0行为键,第3行为值。然后打开文件b,获取第0行中匹配的ID号,以附加匹配的其他列。然后转换回列表并合并两个CSV。
最后添加标题并保存为“output.csv。”
我所期待的例子:
档案a:
1234, 14/12/2,123.4,
5618, 13/1/12,23.4,
9143, 15/2/14,222.4,
文件b:
1234,abc,cda
9143,nda, bad
5618, ede, fpc
(梦)输出:
payment date, payment amount, other id, other other id, payment status, payment type
14/12/2,123.4,1234,abc,cda, Received, Recurring
13/1/12,23.4,9143,nda, bad, Received, Recurring
15/2/14,222.4,5618, ede, fpc,Received, Recurring
(实际)输出:
ID, payment date, payment amount, other id, other other id, payment status, payment type
1234, 14/12/2,123.4,1234,abc,cda, Received, Recurring
5618, 13/1/12,23.4,9143,nda, bad, Received, Recurring
9143, 15/2/14,222.4,5618, ede, fpc,Received, Recurring
代码:
import csv
#create a dict from first csv, with clearing solution id as key
with open("DDS.csv", "rb") as f:
first = {rows[0]: rows[3:] for rows in list(csv.reader(f))}
# compare second csv, append rank, add received recurring columns
with open("report.csv", "rb") as f:
for row in csv.reader(f):
if row and row[0] in first: # row[0] = clearing solution id
first[row[0]].append(row[1]) # row[1] = rank
first[row[0]].append(row[2])
first[row[0]].append('Received')
first[row[0]].append('Recurring')
# convert dict back to list
merged = [(k,) + tuple(v) for k, v in first.items()]
# write list to output csv
with open('output.csv', "w") as f:
writer = csv.DictWriter(f, fieldnames =['ID', 'Payment Date', 'Payment Amount', 'Other ID','other other ID', 'Payment Status', 'Payment Type'])
writer.writeheader()
csv.writer(f).writerows(merged)
奖励积分: 如何从输出CSV中删除第一列?
由于
答案 0 :(得分:0)
你可以摆脱第一个只是不要将(k,)
添加到你的元组并删除
来自字段名称的“ID”。您不需要创建另一个编写器来编写行csv.writer(f).writerows(merged)
?:
merged = [tuple(v) for k, v in first.items()]
with open('output.csv', "w") as f:
writer = csv.DictWriter(f, fieldnames =['Payment Date', 'Payment Amount', 'Other ID','other other ID', 'Payment Status', 'Payment Type'])
writer.writeheader()
writer.writerows(merged)
答案 1 :(得分:-1)
我建议使用经典的python字符串操作代替csv模块。
例如,使用rows.replace(' ','').split(,)[0]
代替rows[0]
可以解决空间问题。