CSV模塊在Python中的應用有些不對勁

Question

我有一個試圖寫入數據具有以下格式的文件的字符串：

100.10 89.7 1,891.43 123.99 

我想轉換為以下格式：

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

但是，使用以下代碼（其中文件filepath在代碼中之前定義為文本文件位置，而data2是字符串：

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

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

我得到以下輸出：

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

我想用引號掩蓋數據中可能出現的逗號，我在做什么錯呢？

Answer 1

您將一個字符串傳遞給writerow ，因此它遍歷字符，在每個字符之間放置一個逗號。 最小修復是：

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

另外，將數據保存在列表中可能比將字符串放在首位更方便！

Answer 2

my_string = '100.10 89.7 1891.43 123.99'

string_list = my_string.split()

然后string_list返回：

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

當您這樣做時：

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

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

它按字符分割data2，因此當您讀回它時，會將每個字符讀為單獨的列。 您可能想要這樣：

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())

編輯

因此，我可能會建議使用其他方法，而忘記注冊方言，這對我而言始終是有問題的。

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()

對我來說：

"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

收益：

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

因此，在編寫時，請執行以下操作：

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

您可能打算寫一堆行，因此請確保遍歷要寫的行，或者使用writerows而不是writerow 。 請參閱文檔中的示例。