在python中重新格式化字典输出

时间:2015-06-05 19:12:35

标签: python csv formatting ordereddictionary

我有一个OrderedDict,我已将它导出到csv,但我希望它的格式不同。

我的阅读,排序和制作字典的代码:

from collections import defaultdict, OrderedDict

counts = defaultdict(lambda: {"User": 0, "Equipment": 0, "Neither": 0})
with open('sorterexample.csv', 'rb') as fh: 
    reader = csv.reader(fh, delimiter=',') 
    headerline = reader.next()
    for row in reader: 
        company, calltype = row[0], row[2]
        counts[company][calltype] += 1
        sorted_counts = OrderedDict(sorted(counts.iteritems(), key=lambda   counts_tup: sum(counts_tup[1].values()), reverse=True))

print(sorted_counts)
    writer = csv.writer(open('sortedcounts.csv', 'wb'))
    for key, value in sorted_counts.items():
        writer.writerow([key, value])

我的产品:

OrderedDict([('Customer1', {'Equipment': 0, 'Neither': 1, 'User': 4}), ('Customer3', {'Equipment': 1, 'Neither': 1, 'User': 2}), ('Customer2', {'Equipment': 0, 'Neither': 0, 'User': 1}), ('Customer4', {'Equipment': 1, 'Neither': 0, 'User': 0})])

我的CSV:

Customer1,  {'Equipment': 0, 'Neither': 1, 'User': 4}
Customer3,  {'Equipment': 1, 'Neither': 1, 'User': 2}
Customer2,  {'Equipment': 0, 'Neither': 0, 'User': 1}
Customer4,  {'Equipment': 1, 'Neither': 0, 'User': 0}

我希望它看起来像这样:

Top Calling Customers,         Equipment,    User,    Neither,
Customer 1,                      0,           4,        1,
Customer 3,                      1,           2,        1,
Customer 2,                      0,           1,        0,
Customer 4,                      1,           0,        0,

我如何格式化它以便在我的csv中以这种方式显示?

编辑:我查看了https://docs.python.org/2.7/howto/sorting.html,itemgetter(),并按照python(How do I sort a list of dictionaries by values of the dictionary in Python?)中的值对字典进行排序,但我仍然无法让它看起来像我想要的那样。 / p>

1 个答案:

答案 0 :(得分:1)

这将按照您描述的方式对其进行格式化:首先,它使用列表中的第一个条目写入标题行,以找出列的名称,然后写入其余行。

writer = csv.writer(open('sortedcounts.csv', 'wb'))
header = ['Top Calling Customers'] + list(list(sorted_counts.values())[0].keys())
writer.writerow(header)
for key, value in sorted_counts.items():
    writer.writerow([key] + list(value.values()))