切片H20框架

问题描述

是否有一种方法可以根据布尔限制对H20框架进行切片？我制作了一个函数，该函数收集True和False'数组。我的目标是基于此布尔数组对H2O帧进行切片。

出于复制目的：

#Creation of Dataset    
X,y = make_classification(n_samples=5000,n_features=15,n_@R_670_4045@ive=15,n_redundant=0,n_repeated=0,n_classes=4,n_clusters_per_class=2,class_sep=3,flip_y=0.1,weights=[0.4,0.20,0.10,0.05],shuffle=True,random_state=1234)
    
    dataset_x = pd.DataFrame({'var1': X[:,0],'var2': X[:,1],'var3': X[:,2]})
    
    dataset_x['var2'] = dataset_x['var3'].round(0)
    
    dataset_x['var3'] = dataset_x['var3']*(-1)
    
    dataset_x['var4'] =np.where(dataset_x['var1']<=0,1)
    
    conditions = [(dataset_x['var2'] <= 0) & (dataset_x['var4'] == 0),(dataset_x['var2'] <=0) & (dataset_x['var4'] == 1),(dataset_x['var2'] >=0) & (dataset_x['var4'] == 0),(dataset_x['var2'] >=0) & (dataset_x['var4'] == 1)]
    
    choices = [0,1,2,3]
    
    dataset_x['var5'] = np.select(conditions,choices,default=0)
    
    dataset_x['var6'] = dataset_x['var3'].abs().round(0)
    
    mean_var1 = dataset_x['var3'].mean()
    len_var1 = len(dataset_x['var3'])
    
    dataset_x['var7'] =(mean_var1*(2.718)**((mean_var1)*(dataset_x['var1'].round(0))*-1))
    
    dataset_x['var8'] =dataset_x['var1'].round(0)
    
    dataset_x['var8'] =abs(dataset_x['var1'].round(0))*2
    
    dataset_y = pd.DataFrame({'target': y})
    
    simulated_irregular_dataset = pd.concat([dataset_x,dataset_y],axis=1)

定义布尔切片器：

def boolean_slicer(size,num_feat):
    array_slicer = []
    for i in range(size):
        slicer = np.ones(num_feat,dtype=np.bool)
        slicer[:int(0.5*num_feat)]=False
        np.random.shuffle(slicer)
        array_slicer.append(slicer)
    return array_slicer

将布尔限制应用于H2O框架：

h2o.init(min_mem_size_GB=8)    
#Transform data into a H2O Frame
H20_df = h2o.H2OFrame(X_train)
print(H20_df)

for i in list_of_Boolean:
    print (i)
    print(H20_df[:,i.tolist()])

收到错误：

H2OResponseError: Server error water.rapids.Rapids.IllegalASTException:
  Error: java.lang.NumberFormatException: For input string: "False"
  Request: POST /99/Rapids
    data: {'ast': '(tmp= py_16_sid_9474 (cols_py Key_Frame__upload_8abf91f6bc0ddcd442f8fa9b6f8b4822.hex [False True True True True True False True]))','session_id': '_sid_9474'}

解决方法

我只需要将布尔值转换为整数即可。

X,y = make_classification(n_samples=5000,n_features=15,n_informative=15,n_redundant=0,n_repeated=0,n_classes=4,n_clusters_per_class=2,class_sep=3,flip_y=0.1,weights=[0.4,0.20,0.10,0.05],shuffle=True,random_state=1234)

dataset_x = pd.DataFrame({'var1': X[:,0],'var2': X[:,1],'var3': X[:,2]})

dataset_x['var2'] = dataset_x['var3'].round(0)

dataset_x['var3'] = dataset_x['var3']*(-1)

dataset_x['var4'] =np.where(dataset_x['var1']<=0,1)

conditions = [(dataset_x['var2'] <= 0) & (dataset_x['var4'] == 0),(dataset_x['var2'] <=0) & (dataset_x['var4'] == 1),(dataset_x['var2'] >=0) & (dataset_x['var4'] == 0),(dataset_x['var2'] >=0) & (dataset_x['var4'] == 1)]

choices = [0,1,2,3]

dataset_x['var5'] = np.select(conditions,choices,default=0)

dataset_x['var6'] = dataset_x['var3'].abs().round(0)

mean_var1 = dataset_x['var3'].mean()
len_var1 = len(dataset_x['var3'])

dataset_x['var7'] =(mean_var1*(2.718)**((mean_var1)*(dataset_x['var1'].round(0))*-1))

dataset_x['var8'] =dataset_x['var1'].round(0)

dataset_x['var8'] =abs(dataset_x['var1'].round(0))*2

dataset_y = pd.DataFrame({'target': y})

simulated_irregular_dataset = pd.concat([dataset_x,dataset_y],axis=1)

def boolean_slicer(size,num_feat):
    array_slicer = []
    for i in range(size):
        slicer = np.ones(num_feat,dtype=np.bool)
        slicer[:int(0.3*num_feat)]=False
        np.random.shuffle(slicer)
        array_slicer.append(slicer)
    return array_slicer

list_of_Boolean = boolean_slicer(20,8)
for i in popo:
    print(i.tolist())

h2o.init(min_mem_size_GB=8)    
#Transform data into a H2O Frame
H20_df = h2o.H2OFrame(X_train)
print(H20_df)

for i in list_of_Boolean:
    print (i)
    i = 1*i
    i = i.astype(int)
    print(H20_df[:,i.tolist()])