如何用python将嵌套字典写入格式整齐的矩阵？

我创建了一个嵌套字典，它表示POS标记符的混淆矩阵，它看起来像：

    file = open(outFile, 'w+')

    matrix = defaultdict(lambda: defaultdict(int))

    for s in range(len(self.goldenTags)):
        for w in range(len(self.goldenTags[s])):
            matrix[self.goldenTags[s][w].tag][self.myTags[s][w].tag] += 1

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在不使用任何库的情况下，仍然可以使用列表创建csv样式的输出

     Tag Tag Tag Tag Tag   
Tag   1   0   2  inf  4
Tag   4   2   0   1   5
Tag  inf inf  1   0   3
Tag   3   4   5   3   0

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使用.xml、.json或.ini代替“重新发明轮子”。大量的图书馆可用于这些和更多。有关简单示例，请使用查看。

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将其放入一个生成字符串的函数中，而不是打印字符串；迭代函数，将返回值写入文件。

什么是“整洁格式”？@PatrickHaugh抱歉。我添加了一个编辑。像这样的东西太好了。你们可能想看看这个问题，寻找一些有趣的选择。

     Tag Tag Tag Tag Tag   
Tag   1   0   2  inf  4
Tag   4   2   0   1   5
Tag  inf inf  1   0   3
Tag   3   4   5   3   0

# create a nested dictionary
d = {'x': {'v1':4, 'v2':5, 'v3':12}, 
     'y':{'v1':2, 'v2':1, 'v3':11}, 
     'z':{'v2':5, 'v3':1}}

# get all of the row and column ids
row_ids = sorted(d.keys())
col_ids = sorted(set(k for v in d.values() for k in v.keys()))

# create an empty list and fill it with the header and then the rows
out = []

# header
out.append(['']+col_ids)

for r in row_ids:
    out.append([r]+[d[r].get(c, 0) for c in col_ids])

out
# returns
[['', 'v1', 'v2', 'v3'], 
 ['x', 4, 5, 12], 
 ['y', 2, 1, 11], 
 ['z', 0, 5, 1]]

d = {'VBP':{'CD': 4,'FW': 1,'JJ': 5,'NN': 61,'NNP': 6,'NNPS': 1,
            'SYM': 2,'VB': 72,'VBD': 5,'VBG': 2,'VBZ': 1},
     'xyz':{'CD': 4,'FW': 1,'JJS': 1,'NN': 61,'NNP': 6,'NNPS': 1,
            'UH': 19,'VB': 72,'VBD': 5,'VBP': 537,'VBZ': 1}}

# find all the columns and all the rows, sort them    
columns = sorted(set(key for dictionary in d.values() for key in dictionary))
rows = sorted(d)

# figure out how wide each column is
col_width = max(max(len(thing) for thing in columns),
                    max(len(thing) for thing in rows)) + 3

# preliminary format string : one column with specific width, right justified
fmt = '{{:>{}}}'.format(col_width)

# format string for all columns plus a 'label' for the row
fmt = fmt * (len(columns) + 1)

# print the header
print(fmt.format('', *columns))

# print the rows
for row in rows:
    dictionary = d[row]
    s = fmt.format(row, *(dictionary.get(col, 'inf') for col in columns))
    print(s)

>>>
            CD     FW     JJ    JJS     NN    NNP   NNPS    SYM     UH     VB    VBD    VBG    VBP    VBZ
    VBP      4      1      5    inf     61      6      1      2    inf     72      5      2    inf      1
    xyz      4      1    inf      1     61      6      1    inf     19     72      5    inf    537      1
>>>