Question

我有一个字符串，我试图写入文件，其中数据具有以下格式：

100.10 89.7 1,891.43 123.99

我希望转换为以下格式：

"100.10","89.7","1,891.43","123.99",

但是使用以下代码（其中filepath在代码中作为文本文件位置定义，data2是一个字符串：

csv.register_dialect('sas', delimiter=',',quoting=csv.QUOTE_ALL)

with open(filepath, "w") as f:
    writer = csv.writer(f, dialect='sas')
    writer.writerow(data2)

我收到了这个输出：

 "1","0","0",".",1,"0","8","9",".","7","1","8","9","1",".","4","3","1","2","3",".","9","9",

我想用引号来掩盖数据中可能存在的逗号，我做错了什么？

Answer 1

您正在将字符串传递给writerow，因此它会对字符进行迭代，并在每个字符之间添加逗号。最小的修复是：

writer.writerow(data2.split(" "))

或者，将数据保存在列表中而不是首先创建字符串可能更方便！

Answer 2

my_string = '100.10 89.7 1891.43 123.99'

string_list = my_string.split()

string_list返回：

['100.10', '89.7', '1891.43', '123.99']

当你这样做时：

csv.register_dialect('sas', delimiter=',',quoting=csv.QUOTE_ALL)

with open(filepath, "w") as f:
    writer = csv.writer(f, dialect='sas')
    writer.writerow(data2)

它按字符拆分data2，所以当你读回它时，它会将每个字符作为一个单独的列读取。你可能想要这个：

csv.register_dialect('sas', delimiter=',',quoting=csv.QUOTE_ALL)

with open(filepath, "w") as f:
    writer = csv.writer(f, dialect='sas')
    writer.writerow(data2.split())

修改

所以我可能会建议一个不同的方法，忘记注册一个方言，对我来说总是有问题。

string_list = ['100.10', '89.7', '1,891.43', '123.99']
with open(filepath, 'w') as f:
    writer = csv.writer(f, quoting=csv.QUOTE_ALL)
    writer.writerow(string_list)

with open(filepath, 'r') as f:
    print f.read()

对我来说打印：

"100.10","89.7","1,891.43","123.99"

和

with open(filepath, 'rU') as f:
    reader = csv.reader(f, quoting=csv.QUOTE_ALL)
    data = next(reader) # next(reader) gives us the iterable's first row.

data

返回：

['100.10', '89.7', '1,891.43', '123.99']

所以当你写作时，请执行以下操作：

with open(filepath, "w") as f:
    writer = csv.writer(f, quoting=csv.QUOTE_ALL)
    writer.writerow(data2.split()) # *write here

您可能打算编写一堆行，因此请务必循环遍历要编写的行，或使用writerows代替writerow。 See the examples in the documentation.

在Python中应用CSV模块不太对劲

2 个答案: