Question

如果我有一个CSV文件，每行有一个字典值（列为[“Location”]，[“MovieDate”]，[“Formatted_Address”]，[“Lat”]，[“Lng”] ），如果我想按Location进行分组，并且追加共享相同MovieDate值的所有Location值，我会被告知使用OrderDict。

ex of data：

Location,MovieDate,Formatted_Address,Lat,Lng
    "Edgebrook Park, Chicago ",Jun-7 A League of Their Own,"Edgebrook Park, 6525 North Hiawatha Avenue, Chicago, IL 60646, USA",41.9998876,-87.7627672
    "Edgebrook Park, Chicago ","Jun-9 It's a Mad, Mad, Mad, Mad World","Edgebrook Park, 6525 North Hiawatha Avenue, Chicago, IL 60646, USA",41.9998876,-87.7627672

对于每个具有相同位置的行（如本示例中的^），我想像这样输出，以便没有重复的位置。

 "Edgebrook Park, Chicago ","Jun-7 A League of Their Own Jun-9 It's a Mad, Mad, Mad, Mad World","Edgebrook Park, 6525 North Hiawatha Avenue, Chicago, IL 60646, USA",41.9998876,-87.7627672

使用ordereddict执行此操作的代码有什么问题？

from collections import OrderedDict

od = OrderedDict()
import csv
with open("MovieDictFormatted.csv") as f,open("MoviesCombined.csv" ,"w") as out:
    r = csv.reader(f)
    wr = csv.writer(out)
    header = next(r)
    for row in r:
        loc,rest = row[0], row[1]
        od.setdefault(loc, []).append(rest)
    wr.writerow(header)
    for loc,vals in od.items():
        wr.writerow([loc]+vals)

我最终得到的是这样的：

['Edgebrook Park, Chicago ', 'Jun-7 A League of Their Own']
['Gage Park, Chicago ', "Jun-9 It's a Mad, Mad, Mad, Mad World"]
['Jefferson Memorial Park, Chicago ', 'Jun-12 Monsters University ', 'Jul-11 Frozen ', 'Aug-8 The Blues Brothers ']
['Commercial Club Playground, Chicago ', 'Jun-12 Despicable Me 2']

问题是我在这种情况下没有显示其他列，我最好怎么做？我还希望将MovieDate值设为一个长字符串，如下所示： 'Jun-12 Monsters University Jul-11 Frozen Aug-8 The Blues Brothers ' 而不是：

'Jun-12 Monsters University ', 'Jul-11 Frozen ', 'Aug-8 The Blues Brothers '

谢谢你们，欣赏它。我是一个蟒蛇诺布。

遗憾的是，将row[0], row[1]更改为row[0], row[1:]并不能满足我的需求。我只想在第二列（MovieDate）中添加值，而不是复制所有其他列。：

['Jefferson Memorial Park, Chicago ', ['Jun-12 Monsters University ', 'Jefferson Memorial Park, 4822 North Long Avenue, Chicago, IL 60630, USA', '41.76083920000001', '-87.6294353'], ['Jul-11 Frozen ', 'Jefferson Memorial Park, 4822 North Long Avenue, Chicago, IL 60630, USA', '41.76083920000001', '-87.6294353'], ['Aug-8 The Blues Brothers ', 'Jefferson Memorial Park, 4822 North Long Avenue, Chicago, IL 60630, USA', '41.76083920000001', '-87.6294353']]

Answer 1

你只需要进行一些更改，你需要加入lat和long，去掉dupe lat和longs我们还需要使用它作为关键：

with open("data.csv") as f,open("new.csv" ,"w") as out:
    r = csv.reader(f)
    wr= csv.writer(out)
    header = next(r)
    for row in r:
        od.setdefault((row[0], row[-2], row[-1]), []).append(" ".join(row[1:-2]))
    wr.writerow(header)
    for loc,vals in od.items():
        wr.writerow([loc[0]] + vals+list(loc[1:]))

输出：

Location,MovieDate,Formatted_Address,Lat,Lng
"Edgebrook Park, Chicago ","Jun-7 A League of Their Own Edgebrook Park, 6525 North Hiawatha Avenue, Chicago, IL 60646, USA","Jun-9 It's a Mad, Mad, Mad, Mad World Edgebrook Park, 6525 North Hiawatha Avenue, Chicago, IL 60646, USA",41.9998876,-87.7627672

A League of Their Own首先是因为它出现在疯狂的疯狂行之前 row[1:-2]获取所有内容，包括纬度，长度和位置，我们将lat和long存储在我们的密钥元组中，以避免在每行末尾重复写入。

使用名称和解压缩可能会让它更容易理解：

with open("data.csv") as f, open("new.csv", "w") as out:
    r = csv.reader(f)
    wr = csv.writer(out)
    header = next(r)
    for row in r:
        loc, mov, form, lat, long = row
        od.setdefault((loc, lat, long), []).append("{} {}".format(mov, form))
    wr.writerow(header)
    for loc, vals in od.items():
        wr.writerow([loc[0]] + vals + list(loc[1:]))

使用csv.Dictwriter保留五列：

od = OrderedDict()
import csv

with open("data.csv") as f, open("new.csv", "w") as out:
    r = csv.DictReader(f,fieldnames=['Location', 'MovieDate', 'Formatted_Address', 'Lat', 'Lng'])
    wr = csv.DictWriter(out, fieldnames=r.fieldnames)
    for row in r:
        od.setdefault(row["Location"], dict(Location=row["Location"], Lat=row["Lat"], Lng=row["Lng"],
                                        MovieDate=[], Formatted_Address=row["Formatted_Address"]))

        od[row["Location"]]["MovieDate"].append(row["MovieDate"])
    for loc, vals in od.items():
        od[loc]["MovieDate"]= ", ".join(od[loc]["MovieDate"])
        wr.writerow(vals)

＃输出：

"Edgebrook Park, Chicago ","Jun-7 A League of Their Own, Jun-9 It's a Mad, Mad, Mad, Mad World","Edgebrook Park, 6525 North Hiawatha Avenue, Chicago, IL 60646, USA",41.9998876,-87.7627672

因此，五列保持不变，我们将"MovieDate"加入到单个字符串中，Formatted_Address=form始终是唯一的，因此我们无需更新。

结果与您想要的相匹配我们需要做的就是连接MovieDate's并删除位置，纬度，Lng和'Formatted_Address'的重复条目。

Answer 2

让我们尝试改变

od.setdefault(loc, []).append(rest)

要

od[loc] = ' '.join([od.get(loc, ''), ' 'join(rest)])

然后保持原样：

wr.writerow([loc]+vals)

Answer 3

假设location是该行的第一项：

dict = {}
for line in f:
    if line[0] not in dict:
        dict[line[0]] = []
    dict[line[0]].append(line[1:])

对于每个位置，您都拥有行的其余部分

for key, value in dict.iteritems():
    out.write(key + value)

python命令dict问题

3 个答案: