水平写入CSV [英] Writing a CSV horizontally

问题描述

假设我们正在从一些来源中读取具有多个键值对的数据。让我们使用下面的列表作为例子：

  [{'key0'：'key0_value0'，'key1'：'key1_value0' }，
 {'key0'：'key0_value1'，'key1'：'key1_value1'}] 
  

从列表中读取第一个项目应该会导致CSV如下所示：

  key_header | 0 
 --------------------------- 
 key0 | key0_value_0 
 key1 | key1_value_0 
  

读取第二个项目现在应该会产生以下结果：

  key_header | 0 | 1 
 ---------------------------------------- 
 key0 | key0_value_0 | key0_value_1 
 key1 | key1_value_0 | key1_value_1 
  

这是水平直到直到。编写此代码的算法超出了我的范围，我不确定 csv模块是否会因为它似乎假设数据将一次写一行。

解决方案

您必须首先收集所有'columns'，然后写入。您可以通过将所有内容转换为列表列表，然后使用 zip（* columns）将列的列表转置为行列表：

  columns = [['key_header'] + sorted（inputlist [0] .keys（））]＃第一列
 
 for i，enumerate（inputlist）中的条目：
 columns.append（[i] + [列[0] [1：]]中的k的入口[k]）
 
 open（outputfilename，'wb'）as output：
 writer = csv.writer（output）
 writer.writerows（zip（* columns））
  pre> 
 
 
显示行输出的演示：
 > >从pprint import pprint 
>>>> inputlist = [{'key0'：'key0_value0'，'key1'：'key1_value0'}，
 ... {'key0'：'key0_value1'，'key1'：'key1_value1'}] 
 >>>> columns = [['key_header'] + sorted（inputlist [0] .keys（））]＃第一列
>>>>对于i，在枚举中输入（inputlist）：
 ... columns.append（[i] + [entries [k] for k in columns [0] [1：]]）
 ... 
>>>> pprint（zip（* columns））
 [（'key_header'，0，1），
（'key0'，'key0_value0'，'key0_value1'），
（'key1' 'key1_value0'，'key1_value1'）] 
  
 
Say we are reading data from some source with multiple key-value pairs. Let's use the following list as an example:
[{'key0': 'key0_value0', 'key1': 'key1_value0'},
 {'key0': 'key0_value1', 'key1': 'key1_value1'}]
Reading the first item from that list should result in a CSV looking like this:
key_header | 0
---------------------------
key0       | key0_value_0
key1       | key1_value_0
Reading the second item should now result in the following:
key_header | 0            | 1
----------------------------------------
key0       | key0_value_0 | key0_value_1
key1       | key1_value_0 | key1_value_1
This goes on horizontally until until. The algorithm to write this is beyond me, and I am not sure if the csv module will work since it appears to assume data will be written a row at a time.
解决方案
You'll have to first collect all your 'columns', then write. You can do that by converting everything to a list of lists, then use zip(*columns) to transpose the list of columns to a list of rows:
columns = [['key_header'] + sorted(inputlist[0].keys())]  # first column

for i, entry in enumerate(inputlist):
    columns.append([i] + [entry[k] for k in columns[0][1:]])

with open(outputfilename, 'wb') as output:
    writer = csv.writer(output)
    writer.writerows(zip(*columns))
Demo showing the row output:
>>> from pprint import pprint
>>> inputlist = [{'key0': 'key0_value0', 'key1': 'key1_value0'},
...  {'key0': 'key0_value1', 'key1': 'key1_value1'}]
>>> columns = [['key_header'] + sorted(inputlist[0].keys())]  # first column
>>> for i, entry in enumerate(inputlist):
...     columns.append([i] + [entry[k] for k in columns[0][1:]])
... 
>>> pprint(zip(*columns))
[('key_header', 0, 1),
 ('key0', 'key0_value0', 'key0_value1'),
 ('key1', 'key1_value0', 'key1_value1')]


                        
这篇关于水平写入CSV的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！