1

我非常接近完成，但无法解决这个问题。我正在写信给 csv，我的代码一直给我这个输出。

 dict,a,b,c,d
 ,,,,
 list,1,2,3,4

我希望它如下所示：

 dict, list
 a,1
 b,2
 c,3
 d,4

代码是：

    ##Opening my dictionary .cvs file
    with open('some_file.csv', mode='r') as infile:
        reader = csv.reader(infile,)
        DICT = {rows[0]:rows[1] for rows in reader if len(rows) == 2}

    ##Opening my enquiry list .cvs file
    datafile = open(self.filename, 'r')
    datareader = csv.reader(datafile)
    n1 = []
    for row in datareader:
        n1.append(row)

        n = list(itertools.chain(*n1))

    headings = ['dict', 'list']

    ##Writing to .cvs file       
    with open(asksaveasfilename(), 'w') as fp:
        a = csv.writer(fp)
        # write row of header names
        a.writerow(n)

        # build up a list of the values in DICT corresponding to the keys in n
        values = []
        for name in n:
            if name in DICT:
                values.append(DICT[name])
            else:
                values.append("Not Available")

        # now write them out
        a.writerow(values)

我尝试使用writerows，但这也会打印错误的数据

d,i,c,t
a
b
c
d
l,i,s,t
1
2
3
4

解决方案：

    for nameValueTuple in zip(n,values):
    a.writerow(nameValueTuple)

做到了

4

2 回答 2

1

直接写入数据

import csv

DICT = {a:a*a for a in [1,2,3,4,5]}
n = [2, 5, 99, 3]

headings = ['dict', 'list']

##Writing to .cvs file       
with open("try.csv", 'w') as fp:
  a = csv.writer(fp)
  a.writerow(headings)
  for name in n:
    if name in DICT:
       a.writerow([name, DICT[name]])
    else:
       a.writerow([name, "Not Available"])

这将导致try.csv包含：

dict,list
2,4
5,25
99,Not Available
3,9

先进行处理，然后写入处理后的行：

您还可以一次完成处理并编写所有内容：

import csv

DICT = {a:a*a for a in [1,2,3,4,5,6]}
ns = [2,3,99,5]

headings = ['dict', 'list']

ns_squared = [DICT[name] if name in DICT else "NOT_FOUND" for name in names]

print(ns_squared) #=> [4, 9, 'NOT_FOUND', 25]

rows = zip(ns,ns_squared)

with open("try.csv", 'w') as fp:
  a = csv.writer(fp)
  a.writerow(headings)
  a.writerows(rows)

这将导致：

dict,list
2,4
3,9
99,NOT_FOUND
5,25

使用 zip 将列变成行

如果您将列作为列表，则可以使用zip()内置函数将它们转换为行。例如：

>>> column1 = ["value", 1, 2, 3, 4]
>>> column2 = ["square", 2, 4, 9, 16]
>>> zip(column1,column2)
[('value', 'square'), (1, 2), (2, 4), (3, 9), (4, 16)]
于 2013-06-25T11:35:27.090 回答
0

自从我完成 python 以来已经很长时间了（我假设变量'values'是一个数组）

我认为你的作家应该是这样的；

#not sure about the syntax, mate...
a.writerow([for x in values])

否则使用bultiin函数zip

希望这可以帮助...

于 2013-06-25T09:33:54.290 回答