我自己写了一段代码。
# Compute confusion matrix
def confusion_matrix(act_labels, pred_labels):
uniqueLabels = list(set(act_labels))
clusters = list(set(pred_labels))
cm = [[0 for i in range(len(clusters))] for i in range(len(uniqueLabels))]
for i, act_label in enumerate(uniqueLabels):
for j, pred_label in enumerate(pred_labels):
if act_labels[j] == act_label:
cm[i][pred_label] = cm[i][pred_label] + 1
return cm
# Example
labels=['a','b','c',
'a','b','c',
'a','b','c',
'a','b','c']
pred=[ 1,1,2,
0,1,2,
1,1,1,
0,1,2]
cnf_matrix = confusion_matrix(labels, pred)
print('\n'.join([''.join(['{:4}'.format(item) for item in row])
for row in cnf_matrix]))
Edit:(Dayyyuumm)刚刚发现我可以轻松地做到这一点熊猫交叉表 https://pandas.pydata.org/pandas-docs/stable/generated/pandas.crosstab.html :-/.
labels=['a','b','c',
'a','b','c',
'a','b','c',
'a','b','c']
pred=[ 1,1,2,
0,1,2,
1,1,1,
0,1,2]
# Create a DataFrame with labels and varieties as columns: df
df = pd.DataFrame({'Labels': labels, 'Clusters': pred})
# Create crosstab: ct
ct = pd.crosstab(df['Labels'], df['Clusters'])
# Display ct
print(ct)