TypeError:“KFold”对象不可迭代
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【中文标题】TypeError:“KFold”对象不可迭代【英文标题】:TypeError: 'KFold' object is not iterable 【发布时间】:2018-07-16 09:56:27 【问题描述】:我正在关注Kaggle 上的一个内核,主要是关注A kernel for Credit Card Fraud Detection。
我已经到了需要执行 KFold 以找到 Logistic 回归的最佳参数的步骤。
以下代码显示在内核本身中,但出于某种原因(可能是旧版本的 scikit-learn,给我一些错误)。
def printing_Kfold_scores(x_train_data,y_train_data):
fold = KFold(len(y_train_data),5,shuffle=False)
# Different C parameters
c_param_range = [0.01,0.1,1,10,100]
results_table = pd.DataFrame(index = range(len(c_param_range),2), columns = ['C_parameter','Mean recall score'])
results_table['C_parameter'] = c_param_range
# the k-fold will give 2 lists: train_indices = indices[0], test_indices = indices[1]
j = 0
for c_param in c_param_range:
print('-------------------------------------------')
print('C parameter: ', c_param)
print('-------------------------------------------')
print('')
recall_accs = []
for iteration, indices in enumerate(fold,start=1):
# Call the logistic regression model with a certain C parameter
lr = LogisticRegression(C = c_param, penalty = 'l1')
# Use the training data to fit the model. In this case, we use the portion of the fold to train the model
# with indices[0]. We then predict on the portion assigned as the 'test cross validation' with indices[1]
lr.fit(x_train_data.iloc[indices[0],:],y_train_data.iloc[indices[0],:].values.ravel())
# Predict values using the test indices in the training data
y_pred_undersample = lr.predict(x_train_data.iloc[indices[1],:].values)
# Calculate the recall score and append it to a list for recall scores representing the current c_parameter
recall_acc = recall_score(y_train_data.iloc[indices[1],:].values,y_pred_undersample)
recall_accs.append(recall_acc)
print('Iteration ', iteration,': recall score = ', recall_acc)
# The mean value of those recall scores is the metric we want to save and get hold of.
results_table.ix[j,'Mean recall score'] = np.mean(recall_accs)
j += 1
print('')
print('Mean recall score ', np.mean(recall_accs))
print('')
best_c = results_table.loc[results_table['Mean recall score'].idxmax()]['C_parameter']
# Finally, we can check which C parameter is the best amongst the chosen.
print('*********************************************************************************')
print('Best model to choose from cross validation is with C parameter = ', best_c)
print('*********************************************************************************')
return best_c
我得到的错误如下:
对于这一行:fold = KFold(len(y_train_data),5,shuffle=False)
错误:
TypeError: init() 为参数 'shuffle' 获得了多个值
如果我从这一行中删除 shuffle=False
,我会收到以下错误:
TypeError: shuffle must be True or False;得到 5 个
如果我删除 5
并保留 shuffle=False
,我会收到以下错误;
TypeError: 'KFold' 对象不可迭代 来自这一行:
for iteration, indices in enumerate(fold,start=1):
如果有人可以帮助我解决这个问题并建议如何使用最新版本的 scikit-learn 来完成,我们将不胜感激。
谢谢。
【问题讨论】:
【参考方案1】:这取决于您如何导入 KFold。
如果你这样做了:
from sklearn.cross_validation import KFold
那么你的代码应该可以工作了。因为它需要 3 个参数:- 数组长度、分割数和随机播放
但如果你这样做:
from sklearn.model_selection import KFold
那么这将不起作用,您只需要传递拆分和随机播放的数量。在enumerate()
中进行更改时无需传递数组的长度。
顺便说一下,model_selection 是新模块,推荐使用。尝试像这样使用它:
fold = KFold(5,shuffle=False)
for train_index, test_index in fold.split(X):
# Call the logistic regression model with a certain C parameter
lr = LogisticRegression(C = c_param, penalty = 'l1')
# Use the training data to fit the model. In this case, we use the portion of the fold to train the model
lr.fit(x_train_data.iloc[train_index,:], y_train_data.iloc[train_index,:].values.ravel())
# Predict values using the test indices in the training data
y_pred_undersample = lr.predict(x_train_data.iloc[test_index,:].values)
# Calculate the recall score and append it to a list for recall scores representing the current c_parameter
recall_acc = recall_score(y_train_data.iloc[test_index,:].values,y_pred_undersample)
recall_accs.append(recall_acc)
【讨论】:
感谢Vivek提供的额外信息,我确实在使用model_selection,因此不明白导致此错误的原因,现在我知道了,在您为我澄清后,谢谢。【参考方案2】:KFold 是一个拆分器,所以你必须给一些东西来拆分。
示例代码:
X = np.array([1,1,1,1], [2,2,2,2], [3,3,3,3], [4,4,4,4]])
y = np.array([1, 2, 3, 4])
# Now you create your Kfolds by the way you just have to pass number of splits and if you want to shuffle.
fold = KFold(2,shuffle=False)
# For iterate over the folds just use split
for train_index, test_index in fold.split(X):
X_train, X_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
# Follow fitting the classifier
如果你想获取训练/测试循环的索引,只需添加枚举
for i, train_index, test_index in enumerate(fold.split(X)):
print('Iteration:', i)
X_train, X_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
我希望这行得通
【讨论】:
嘿 Tzomas,谢谢你的回答,它确实解决了错误问题,但我真的不明白为什么要拆分 X?在内核本身中,它对每个参数 C 迭代 5 次,但在我的情况下,它迭代 X 的长度数,远远高于 5,这里有什么问题? 对不起,我想我的错字有一些错误,它确实解决了我的问题,谢谢!以上是关于TypeError:“KFold”对象不可迭代的主要内容,如果未能解决你的问题,请参考以下文章
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