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请注意:我为了这项任务而与 numpy 结婚。
我正在尝试编写一个函数来实现以下目标:
例如,如果我有 input=100 个样本(行)
,并且有两个类(由最后一列中的值表示),A 和 B,分别为 33% 和 67 % split,然后我应该创建 5 个折叠,每个折叠包含 20 个样本,其中 6 或 7 个样本是 A,13 或 14 个样本是 B。
这就是我正在努力实现的目标。我不知道如何正确确保 FOLD 本身包含类的正确采样分布。
我有以下代码来显示我迄今为止的尝试。到目前为止,我已经编写了两个函数,它们能够告诉我输入类的分布是什么,并且能够创建 5 个折叠。然而,我需要找到一种方法来组合这些并创建 5 个折叠来维持各自的分布。
import numpy
def csv_to_array(file):
# Open the file, and load it in delimiting on the ',' for a comma separated value file
data = open(file, 'r')
data = numpy.loadtxt(data, delimiter=',')
# Loop through the data in the array
for index in range(len(data)):
# Utilize a try catch to try and convert to float, if it can't convert to float, converts to 0
try:
data[index] = [float(x) for x in data[index]]
except Exception:
data[index] = 0
except ValueError:
data[index] = 0
# Return the now type-formatted data
return data
def class_distribution(dataset):
dataset = numpy.asarray(dataset)
num_total_rows = dataset.shape[0]
num_columns = dataset.shape[1]
classes = dataset[:,num_columns-1]
classes = numpy.unique(classes)
for aclass in classes:
total = 0
for row in dataset:
if numpy.array_equal(aclass, row[-1]):
total = total + 1
else:
continue
print(aclass, " Has: ", ((total/num_total_rows) * 100))
print(aclass, " : ", total)
def create_folds(dataset):
# print("DATASET", dataset)
numpy.random.shuffle(dataset)
num_rows = dataset.shape[0]
split_mark = int(num_rows / 5)
folds = []
fold_sets = []
temp1 = dataset[:split_mark]
# print("TEMP1", temp1)
temp2 = dataset[split_mark:split_mark*2]
# print("TEMP2", temp2)
temp3 = dataset[split_mark*2:split_mark*3]
# print("TEMP3", temp3)
temp4 = dataset[split_mark*3:split_mark*4]
# print("TEMP4", temp4)
temp5 = dataset[split_mark*4:]
# print("TEMP5", temp5)
folds.append(temp1)
folds.append(temp2)
folds.append(temp3)
folds.append(temp4)
folds.append(temp5)
folds = numpy.asarray(folds)
# print(folds)
return folds
def main():
print("BEGINNING CFV")
ecoli = csv_to_array('Classification/ecoli.csv')
# print(len(ecoli))
class_distribution(ecoli)
create_folds(ecoli)
main()
这是我正在使用的 csv 示例,最后一列表示类。它是 ecoli dataset 的修改来自 UCI 机器学习存储库:
0.61,0.45,0.48,0.5,0.48,0.35,0.41,0
0.17,0.38,0.48,0.5,0.45,0.42,0.5,0
0.44,0.35,0.48,0.5,0.55,0.55,0.61,0
0.43,0.4,0.48,0.5,0.39,0.28,0.39,0
0.42,0.35,0.48,0.5,0.58,0.15,0.27,0
0.23,0.33,0.48,0.5,0.43,0.33,0.43,0
0.37,0.52,0.48,0.5,0.42,0.42,0.36,0
0.29,0.3,0.48,0.5,0.45,0.03,0.17,0
0.22,0.36,0.48,0.5,0.35,0.39,0.47,0
0.23,0.58,0.48,0.5,0.37,0.53,0.59,0
0.47,0.47,0.48,0.5,0.22,0.16,0.26,0
0.54,0.47,0.48,0.5,0.28,0.33,0.42,0
0.51,0.37,0.48,0.5,0.35,0.36,0.45,0
0.4,0.35,0.48,0.5,0.45,0.33,0.42,0
0.44,0.34,0.48,0.5,0.3,0.33,0.43,0
0.44,0.49,0.48,0.5,0.39,0.38,0.4,0
0.43,0.32,0.48,0.5,0.33,0.45,0.52,0
0.49,0.43,0.48,0.5,0.49,0.3,0.4,0
0.47,0.28,0.48,0.5,0.56,0.2,0.25,0
0.32,0.33,0.48,0.5,0.6,0.06,0.2,0
0.34,0.35,0.48,0.5,0.51,0.49,0.56,0
0.35,0.34,0.48,0.5,0.46,0.3,0.27,0
0.38,0.3,0.48,0.5,0.43,0.29,0.39,0
0.38,0.44,0.48,0.5,0.43,0.2,0.31,0
0.41,0.51,0.48,0.5,0.58,0.2,0.31,0
0.34,0.42,0.48,0.5,0.41,0.34,0.43,0
0.51,0.49,0.48,0.5,0.53,0.14,0.26,0
0.25,0.51,0.48,0.5,0.37,0.42,0.5,0
0.29,0.28,0.48,0.5,0.5,0.42,0.5,0
0.25,0.26,0.48,0.5,0.39,0.32,0.42,0
0.24,0.41,0.48,0.5,0.49,0.23,0.34,0
0.17,0.39,0.48,0.5,0.53,0.3,0.39,0
0.04,0.31,0.48,0.5,0.41,0.29,0.39,0
0.61,0.36,0.48,0.5,0.49,0.35,0.44,0
0.34,0.51,0.48,0.5,0.44,0.37,0.46,0
0.28,0.33,0.48,0.5,0.45,0.22,0.33,0
0.4,0.46,0.48,0.5,0.42,0.35,0.44,0
0.23,0.34,0.48,0.5,0.43,0.26,0.37,0
0.37,0.44,0.48,0.5,0.42,0.39,0.47,0
0,0.38,0.48,0.5,0.42,0.48,0.55,0
0.39,0.31,0.48,0.5,0.38,0.34,0.43,0
0.3,0.44,0.48,0.5,0.49,0.22,0.33,0
0.27,0.3,0.48,0.5,0.71,0.28,0.39,0
0.17,0.52,0.48,0.5,0.49,0.37,0.46,0
0.36,0.42,0.48,0.5,0.53,0.32,0.41,0
0.3,0.37,0.48,0.5,0.43,0.18,0.3,0
0.26,0.4,0.48,0.5,0.36,0.26,0.37,0
0.4,0.41,0.48,0.5,0.55,0.22,0.33,0
0.22,0.34,0.48,0.5,0.42,0.29,0.39,0
0.44,0.35,0.48,0.5,0.44,0.52,0.59,0
0.27,0.42,0.48,0.5,0.37,0.38,0.43,0
0.16,0.43,0.48,0.5,0.54,0.27,0.37,0
0.06,0.61,0.48,0.5,0.49,0.92,0.37,1
0.44,0.52,0.48,0.5,0.43,0.47,0.54,1
0.63,0.47,0.48,0.5,0.51,0.82,0.84,1
0.23,0.48,0.48,0.5,0.59,0.88,0.89,1
0.34,0.49,0.48,0.5,0.58,0.85,0.8,1
0.43,0.4,0.48,0.5,0.58,0.75,0.78,1
0.46,0.61,0.48,0.5,0.48,0.86,0.87,1
0.27,0.35,0.48,0.5,0.51,0.77,0.79,1
0.52,0.39,0.48,0.5,0.65,0.71,0.73,1
0.29,0.47,0.48,0.5,0.71,0.65,0.69,1
0.55,0.47,0.48,0.5,0.57,0.78,0.8,1
0.12,0.67,0.48,0.5,0.74,0.58,0.63,1
0.4,0.5,0.48,0.5,0.65,0.82,0.84,1
0.73,0.36,0.48,0.5,0.53,0.91,0.92,1
0.84,0.44,0.48,0.5,0.48,0.71,0.74,1
0.48,0.45,0.48,0.5,0.6,0.78,0.8,1
0.54,0.49,0.48,0.5,0.4,0.87,0.88,1
0.48,0.41,0.48,0.5,0.51,0.9,0.88,1
0.5,0.66,0.48,0.5,0.31,0.92,0.92,1
0.72,0.46,0.48,0.5,0.51,0.66,0.7,1
0.47,0.55,0.48,0.5,0.58,0.71,0.75,1
0.33,0.56,0.48,0.5,0.33,0.78,0.8,1
0.64,0.58,0.48,0.5,0.48,0.78,0.73,1
0.11,0.5,0.48,0.5,0.58,0.72,0.68,1
0.31,0.36,0.48,0.5,0.58,0.94,0.94,1
0.68,0.51,0.48,0.5,0.71,0.75,0.78,1
0.69,0.39,0.48,0.5,0.57,0.76,0.79,1
0.52,0.54,0.48,0.5,0.62,0.76,0.79,1
0.46,0.59,0.48,0.5,0.36,0.76,0.23,1
0.36,0.45,0.48,0.5,0.38,0.79,0.17,1
0,0.51,0.48,0.5,0.35,0.67,0.44,1
0.1,0.49,0.48,0.5,0.41,0.67,0.21,1
0.3,0.51,0.48,0.5,0.42,0.61,0.34,1
0.61,0.47,0.48,0.5,0,0.8,0.32,1
0.63,0.75,0.48,0.5,0.64,0.73,0.66,1
0.71,0.52,0.48,0.5,0.64,1,0.99,1
0.72,0.42,0.48,0.5,0.65,0.77,0.79,2
0.79,0.41,0.48,0.5,0.66,0.81,0.83,2
0.83,0.48,0.48,0.5,0.65,0.76,0.79,2
0.69,0.43,0.48,0.5,0.59,0.74,0.77,2
0.79,0.36,0.48,0.5,0.46,0.82,0.7,2
0.78,0.33,0.48,0.5,0.57,0.77,0.79,2
0.75,0.37,0.48,0.5,0.64,0.7,0.74,2
0.59,0.29,0.48,0.5,0.64,0.75,0.77,2
0.67,0.37,0.48,0.5,0.54,0.64,0.68,2
0.66,0.48,0.48,0.5,0.54,0.7,0.74,2
0.64,0.46,0.48,0.5,0.48,0.73,0.76,2
0.76,0.71,0.48,0.5,0.5,0.71,0.75,2
0.84,0.49,0.48,0.5,0.55,0.78,0.74,2
0.77,0.55,0.48,0.5,0.51,0.78,0.74,2
0.81,0.44,0.48,0.5,0.42,0.67,0.68,2
0.58,0.6,0.48,0.5,0.59,0.73,0.76,2
0.63,0.42,0.48,0.5,0.48,0.77,0.8,2
0.62,0.42,0.48,0.5,0.58,0.79,0.81,2
0.86,0.39,0.48,0.5,0.59,0.89,0.9,2
0.81,0.53,0.48,0.5,0.57,0.87,0.88,2
0.87,0.49,0.48,0.5,0.61,0.76,0.79,2
0.47,0.46,0.48,0.5,0.62,0.74,0.77,2
0.76,0.41,0.48,0.5,0.5,0.59,0.62,2
0.7,0.53,0.48,0.5,0.7,0.86,0.87,2
0.64,0.45,0.48,0.5,0.67,0.61,0.66,2
0.81,0.52,0.48,0.5,0.57,0.78,0.8,2
0.73,0.26,0.48,0.5,0.57,0.75,0.78,2
0.49,0.61,1,0.5,0.56,0.71,0.74,2
0.88,0.42,0.48,0.5,0.52,0.73,0.75,2
0.84,0.54,0.48,0.5,0.75,0.92,0.7,2
0.63,0.51,0.48,0.5,0.64,0.72,0.76,2
0.86,0.55,0.48,0.5,0.63,0.81,0.83,2
0.79,0.54,0.48,0.5,0.5,0.66,0.68,2
0.57,0.38,0.48,0.5,0.06,0.49,0.33,2
0.78,0.44,0.48,0.5,0.45,0.73,0.68,2
0.78,0.68,0.48,0.5,0.83,0.4,0.29,3
0.63,0.69,0.48,0.5,0.65,0.41,0.28,3
0.67,0.88,0.48,0.5,0.73,0.5,0.25,3
0.61,0.75,0.48,0.5,0.51,0.33,0.33,3
0.67,0.84,0.48,0.5,0.74,0.54,0.37,3
0.74,0.9,0.48,0.5,0.57,0.53,0.29,3
0.73,0.84,0.48,0.5,0.86,0.58,0.29,3
0.75,0.76,0.48,0.5,0.83,0.57,0.3,3
0.77,0.57,0.48,0.5,0.88,0.53,0.2,3
0.74,0.78,0.48,0.5,0.75,0.54,0.15,3
0.68,0.76,0.48,0.5,0.84,0.45,0.27,3
0.56,0.68,0.48,0.5,0.77,0.36,0.45,3
0.65,0.51,0.48,0.5,0.66,0.54,0.33,3
0.52,0.81,0.48,0.5,0.72,0.38,0.38,3
0.64,0.57,0.48,0.5,0.7,0.33,0.26,3
0.6,0.76,1,0.5,0.77,0.59,0.52,3
0.69,0.59,0.48,0.5,0.77,0.39,0.21,3
0.63,0.49,0.48,0.5,0.79,0.45,0.28,3
0.71,0.71,0.48,0.5,0.68,0.43,0.36,3
0.68,0.63,0.48,0.5,0.73,0.4,0.3,3
0.74,0.49,0.48,0.5,0.42,0.54,0.36,4
0.7,0.61,0.48,0.5,0.56,0.52,0.43,4
0.66,0.86,0.48,0.5,0.34,0.41,0.36,4
0.73,0.78,0.48,0.5,0.58,0.51,0.31,4
0.65,0.57,0.48,0.5,0.47,0.47,0.51,4
0.72,0.86,0.48,0.5,0.17,0.55,0.21,4
0.67,0.7,0.48,0.5,0.46,0.45,0.33,4
0.67,0.81,0.48,0.5,0.54,0.49,0.23,4
0.67,0.61,0.48,0.5,0.51,0.37,0.38,4
0.63,1,0.48,0.5,0.35,0.51,0.49,4
0.57,0.59,0.48,0.5,0.39,0.47,0.33,4
0.71,0.71,0.48,0.5,0.4,0.54,0.39,4
0.66,0.74,0.48,0.5,0.31,0.38,0.43,4
0.67,0.81,0.48,0.5,0.25,0.42,0.25,4
0.64,0.72,0.48,0.5,0.49,0.42,0.19,4
0.68,0.82,0.48,0.5,0.38,0.65,0.56,4
0.32,0.39,0.48,0.5,0.53,0.28,0.38,4
0.7,0.64,0.48,0.5,0.47,0.51,0.47,4
0.63,0.57,0.48,0.5,0.49,0.7,0.2,4
0.69,0.65,0.48,0.5,0.63,0.48,0.41,4
0.43,0.59,0.48,0.5,0.52,0.49,0.56,4
0.74,0.56,0.48,0.5,0.47,0.68,0.3,4
0.71,0.57,0.48,0.5,0.48,0.35,0.32,4
0.61,0.6,0.48,0.5,0.44,0.39,0.38,4
0.59,0.61,0.48,0.5,0.42,0.42,0.37,4
0.74,0.74,0.48,0.5,0.31,0.53,0.52,4
最佳答案
在听取了@AlexL的建议后,我查看了StratifiedKFold代码并开发了具有以下两个功能的修改版本:
# This function returns the list of classes, and their associated weights (i.e. distributions)
def class_distribution(dataset):
dataset = numpy.asarray(dataset)
num_total_rows = dataset.shape[0]
num_columns = dataset.shape[1]
classes = dataset[:, num_columns - 1]
classes = numpy.unique(classes)
class_weights = []
# Loop through the classes one by one
for aclass in classes:
total = 0
weight = 0
for row in dataset:
if numpy.array_equal(aclass, row[-1]):
total = total + 1
else:
continue
weight = float((total / num_total_rows))
class_weights.append(weight)
class_weights = numpy.asarray(class_weights)
return classes, class_weights
# This functions performs k cross fold validation for classification
def cross_fold_validation_classification(dataset, k):
temp_dataset = numpy.asarray(dataset)
classes, class_weights = class_distribution(temp_dataset)
total_num_rows = temp_dataset.shape[0]
data = numpy.copy(temp_dataset)
total_fold_array = []
for _ in range(k):
curr_fold_array = []
# Loop through each class and its associated weight
for a_class, a_class_weight in zip(classes, class_weights):
numpy.random.shuffle(data)
num_added = 0
num_to_add = float((((a_class_weight * total_num_rows)) / k))
tot = 0
for row in data:
curr = row[-1]
if num_added >= num_to_add:
break
else:
if (a_class == curr):
curr_fold_array.append(row)
num_added = num_added + 1
numpy.delete(data, tot)
tot = tot + 1
total_fold_array.append(curr_fold_array)
return total_fold_array
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尝试为基于文本的多标签分类问题训练单层神经网络。 model= Sequential() model.add(Dense(20, input_dim=400, kernel_initializer='
首先是一个简单的例子 import numpy as np a = np.ones((2,2)) b = 2*np.ones((2,2)) c = 3*np.ones((2,2)) d = 4*np.
我正在尝试平均二维 numpy 数组。所以,我使用了 numpy.mean 但结果是空数组。 import numpy as np ws1 = np.array(ws1) ws1_I8 = np.ar
import numpy as np x = np.array([[1,2 ,3], [9,8,7]]) y = np.array([[2,1 ,0], [1,0,2]]) x[y] 预期输出: ar
我有两个数组 A (4000,4000),其中只有对角线填充了数据,而 B (4000,5) 填充了数据。有没有比 numpy.dot(a,b) 函数更快的方法来乘(点)这些数组? 到目前为止,我发现
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