Issue
I have been trying to implement this code in python 2.7. It gives me this error. I would appreciate help. I have latest version of sklearn(0.18.1) and xgboost(0.6)
import xgboost as xgb
from sklearn.model_selection import StratifiedKFold
from sklearn.metrics import f1_score, roc_auc_score, confusion_matrix
nfold = 3
kf = StratifiedKFold(nfold, shuffle=True)
dtrain = xgb.DMatrix(x_train, label=y_train)
dtest = xgb.DMatrix(x_test)
params = {
'objective' : 'binary:logistic',
'eval_metric': 'auc',
'min_child_weight':10,
'scale_pos_weight':scale,
}
hist = xgb.cv(params, dtrain, num_boost_round=10000, folds=kf, early_stopping_rounds=50, as_pandas=True, verbose_eval=100, show_stdv=True, seed=0)
I get this error:
TypeErrorTraceback (most recent call last)
<ipython-input-52-41c415e116d7> in <module>()
5 'scale_pos_weight':scale,
6 }
----> 7 hist = xgb.cv(params, dtrain, num_boost_round=10000, folds=kf, early_stopping_rounds=50, as_pandas=True, verbose_eval=100, show_stdv=True, seed=0)
8
9
/opt/conda/lib/python2.7/site-packages/xgboost/training.pyc in cv(params, dtrain, num_boost_round, nfold, stratified, folds, metrics, obj, feval, maximize, early_stopping_rounds, fpreproc, as_pandas, verbose_eval, show_stdv, seed, callbacks)
369
370 results = {}
--> 371 cvfolds = mknfold(dtrain, nfold, params, seed, metrics, fpreproc, stratified, folds)
372
373 # setup callbacks
/opt/conda/lib/python2.7/site-packages/xgboost/training.pyc in mknfold(dall, nfold, param, seed, evals, fpreproc, stratified, folds)
236 idset = [randidx[(i * kstep): min(len(randidx), (i + 1) * kstep)] for i in range(nfold)]
237 elif folds is not None:
--> 238 idset = [x[1] for x in folds]
239 nfold = len(idset)
240 else:
TypeError: 'StratifiedKFold' object is not iterable
Solution
Inside the xgb.cv
function, try to replace
folds=kf
with
folds=list(kf.split(x_train,y_train))
The split method is applied in order to get the split into training and validation. We then convert it into a list
so that it would be an iterable object.
If that doesn't work, try without the list
. That is:
folds=kf.split(x_train,y_train)
Answered By - Miriam Farber
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