Question

我有一堆列表，我已压缩：

 zipped = zip(list1, list2, list3)

我有一个csv文件（已经有标题）看起来像这样：

Col_1    Col_2    Col_3
data     data     data

现在我想将我的压缩列表添加为已存在文件的列，以便它看起来像这样（列表名称为标题）：

Col_1    Col_2    Col_3    list1    list2    list3
data     data     data     data     data     data

我现在有了这个代码：

with open('existing.csv', 'rb') as infile, open('out_file.csv', 'wb') as outfile:
    reader = csv.reader(infile)
    writer = csv.writer(outfile)
    writer.writerow(["list1", "list2", "list3"])
    for i, row in enumerate(reader):
        row.extend(zipped[i])
        writer.writerow(row)

但是，不幸的是，这会覆盖现有的列名。

Answer 1

您需要保留现有标题，即读者对象的第一行：

with open('existing.csv', 'rb') as infile, open('out_file.csv', 'wb') as outfile:
    reader = csv.reader(infile)
    header = next(reader) # get header
    writer = csv.writer(outfile)
    # create list with old header and new
    writer.writerow(header+["list1", "list2", "list3"])

您还可以使用itertools.izip压缩for循环中调用next的列表来获取每个子元素：

from itertools import izip
zipped = izip(list1, list2, list3)
with open('existing.csv', 'rb') as infile, open('out_file.csv', 'wb') as outfile:
    reader = csv.reader(infile)
    headers = next(reader)
    writer = csv.writer(outfile)
    writer.writerow(headers+["list1", "list2", "list3"])
    for  row in  reader:
        row.extend(next(zipped))
        writer.writerow(row)

你自己的代码实际上不会覆盖标题，它会添加你的新标题，然后直接用压缩列表中的第一个子元素压缩原始标题。

将标头添加到csv文件

1 个答案: