[英]Is it possible to make a confusion matrix from character pairs?
在光学字符识别的背景下,我将尽力总结我的问题:
我有参考句和预测句。
使用Levenshtein editops function ,我制作了一个列表,其中包含一个元组,其中包含:步骤类型(插入、替换、替换)、在参考序列中修改的字符、在预测序列中修改的字符,最后是这些更改的次数在所有参考句子中都进行了(实际上,这些错误对返回的最大出现次数)
[(('insert', 'e', 'm'), 11), (('insert', 't', 'a'), 8), (('insert', 'r', 'o'), 5), (('replace', 'a', 'e'), 2), (('replace', 't', 'T'), 1), (('replace', 'r', 'R'), 1), (('replace', 'M', 'm'), 1), (('delete', ' ', 'a'), 1), (('replace', 'p', 'o'), 1), (('replace', 't', 'a'), 1), (('replace', 'e', 'e'), 1), (('replace', ' ', 'r'), 1), (('insert', ' ', 'd'), 1), (('replace', ' ', 'd'), 1), (('replace', 'i', 'e'), 1), (('replace', 'l', 's'), 1)]
Output 示例
Predicted e m t a r ...continue
Reference
e 1 11 0 0 0
m 0 0 0 0 0
t 0 0 0 8 0
a 2 0 0 0 0
r 0 0 0 0 0
...continue
或像这样(没有标签):
[[1 11 0 0 0
0 0 0 0 0
0 0 0 8 0
2 0 0 0 0
0 0 0 0 0]]
注意:当未遇到字符错误对时,此“矩阵”示例中的默认值 0 将被替换。
一个轨道来解决它? 提前致谢。
我会用计数器做到这一点:
operations = [(('insert', 'e', 'm'), 11), (('insert', 't', 'a'), 8), (('insert', 'r', 'o'), 5), (('replace', 'a', 'e'), 2), (('replace', 't', 'T'), 1), (('replace', 'r', 'R'), 1), (('replace', 'M', 'm'), 1), (('delete', ' ', 'a'), 1), (('replace', 'p', 'o'), 1), (('replace', 't', 'a'), 1), (('replace', 'e', 'e'), 1), (('replace', ' ', 'r'), 1), (('insert', ' ', 'd'), 1), (('replace', ' ', 'd'), 1), (('replace', 'i', 'e'), 1), (('replace', 'l', 's'), 1)]
from collections import defaultdict, Counter
intermediary = defaultdict(Counter)
for (_, src, tgt), count in source:
intermediary[src][tgt] = count
letters = sorted({key for inner in intermediary.values() for key in inner} | set(intermediary.keys()))
confusion_matrix = [[intermediary[src][tgt] for tgt in letters] for src in letters]
结果如下所示:
[[0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 11, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
对于绘图,请参阅此问题的答案:
import seaborn as sn
import pandas as pd
import matplotlib.pyplot as plt
df_cm = pd.DataFrame(confusion_matrix, letters, letters)
sn.set(font_scale=1.4) # for label size
sn.heatmap(df_cm, annot=True, annot_kws={"size": 16}) # font size
plt.show()
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