过样本二进制[重复]

Question

这个问题在这里已有答案：

• Duplicating training examples to handle class imbalance in a pandas data frame 1回答

我有一些由类（X）和一些二进制（Y）组成的数据。我想通过对较小的类进行过采样来均衡类的大小。例如，如果我开始：

Df_01 = pd.DataFrame({'X' : [1,1,1,1,1,1,1,2,2],
                      'Y1': [1,1,1,1,1,0,0,0,1],
                      'Y2': [0,0,0,0,0,1,0,0,0]})

然后我想得到：

Df_02 = pd.DataFrame({'X' : [1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2],
                      'Y1': [1,1,1,1,1,0,0,0,1,0,1,0,1,0,1,0],
                      'Y2': [0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0]})

我试图这样做：

# Sort the data by class
Ma_01 = Df_01.groupby('X')
Di_01 = {}
for name, group in Ma_01:
    Di_01[str(name)] = group

# Size of each class
Se_01 = Df_01.groupby('X').size()

# Size of the biggest class
In_Bi = max(Se_01)

# How much over sampling would equalise the class sizes?
Se_Ra =  In_Bi / Se_01
Di_Ra =  Se_Ra.to_dict()

但是当我尝试：

# Copy each dataframe
Di_03 = {}
for x in Di_01:
    for y in range(int(Di_Ra[int(x)])):
        if not Di_03:
            Di_03[x] = Di_01[x]
        else:
            Di_03[x] = Di_03[x] .append(Di_01[x])

# Concatonate the dictionary to a single dataframe
df_03 = pd.concat(Di_03.values(), ignore_index=True)

我明白了

KeyError: '2'

Answer 1

感谢您找到重复的Matthew Strawbridge！ Ayhan对我的数据的原始作品的答案：

max_size = Df_01['X'].value_counts().max()

lst = [Df_01]

for class_index, group in Df_01.groupby('X'):
    lst.append(group.sample(max_size-len(group), replace=True))

Df_03 = pd.concat(lst)