Dat*_*Man 5 python keras tensorflow loss-function pandas-groupby
我正在编写一个自定义损失函数,需要计算每组预测值的比率。作为一个简化的例子,这里是我的数据和模型代码的样子:
def main():
df = pd.DataFrame(columns=["feature_1", "feature_2", "condition_1", "condition_2", "label"],
data=[[5, 10, "a", "1", 0],
[30, 20, "a", "1", 1],
[50, 40, "a", "1", 0],
[15, 20, "a", "2", 0],
[25, 30, "b", "2", 1],
[35, 40, "b", "1", 0],
[10, 80, "b", "1", 1]])
features = ["feature_1", "feature_2"]
conds_and_label = ["condition_1", "condition_2", "label"]
X = df[features]
Y = df[conds_and_label]
model = my_model(input_shape=len(features))
model.fit(X, Y, epochs=10, batch_size=128)
model.evaluate(X, Y)
def custom_loss(conditions, y_pred): # this is what I need help with
conds = ["condition_1", "condition_2"]
conditions["label_pred"] = y_pred
g = conditions.groupby(by=conds,
as_index=False).apply(lambda x: x["label_pred"].sum() /
len(x)).reset_index(name="pred_ratio")
# true_ratios will be a constant, external DataFrame. Simplified example here:
true_ratios = pd.DataFrame(columns=["condition_1", "condition_2", "true_ratio"],
data=[["a", "1", 0.1],
["a", "2", 0.2],
["b", "1", 0.8],
["b", "2", 0.9]])
merged = pd.merge(g, true_ratios, on=conds)
merged["diff"] = merged["pred_ratio"] - merged["true_ratio"]
return K.mean(K.abs(merged["diff"]))
def joint_loss(conds_and_label, y_pred):
y_true = conds_and_label[:, 2]
conditions = tf.gather(conds_and_label, [0, 1], axis=1)
loss_1 = standard_loss(y_true=y_true, y_pred=y_pred) # not shown
loss_2 = custom_loss(conditions=conditions, y_pred=y_pred)
return 0.5 * loss_1 + 0.5 * loss_2
def my_model(input_shape=None):
model = Sequential()
model.add(Dense(units=2, activation="relu"), input_shape=(input_shape,))
model.add(Dense(units=1, activation='sigmoid'))
model.add(Flatten())
model.compile(loss=joint_loss, optimizer="Adam",
metrics=[joint_loss, custom_loss, "accuracy"])
return model
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我需要帮助的是custom_loss函数。如您所见,它目前被编写为好像输入是 Pandas DataFrames。但是,输入将是 Keras Tensors(带有 tensorflow 后端),因此我试图弄清楚如何将当前代码转换custom_loss为使用 Keras/TF 后端函数。例如,我在网上搜索并找不到在 Keras/TF 中进行 groupby 以获得我需要的比率的方法......
一些可能对您有帮助的上下文/解释:
joint_loss,它由standard_loss(未显示)和 组成custom_loss。但我只需要帮助转换custom_loss.custom_loss是:
我最终找到了解决方案,尽管我想要一些反馈(特别是某些部分)。这是解决方案:
import pandas as pd
import tensorflow as tf
import keras.backend as K
from keras.models import Sequential
from keras.layers import Dense, Flatten, Dropout
from tensorflow.python.ops import gen_array_ops
def main():
df = pd.DataFrame(columns=["feature_1", "feature_2", "condition_1", "condition_2", "label"],
data=[[5, 10, "a", "1", 0],
[30, 20, "a", "1", 1],
[50, 40, "a", "1", 0],
[15, 20, "a", "2", 0],
[25, 30, "b", "2", 1],
[35, 40, "b", "1", 0],
[10, 80, "b", "1", 1]])
df = pd.concat([df] * 500) # making data artificially larger
true_ratios = pd.DataFrame(columns=["condition_1", "condition_2", "true_ratio"],
data=[["a", "1", 0.1],
["a", "2", 0.2],
["b", "1", 0.8],
["b", "2", 0.9]])
features = ["feature_1", "feature_2"]
conditions = ["condition_1", "condition_2"]
conds_ratios_label = conditions + ["true_ratio", "label"]
df = pd.merge(df, true_ratios, on=conditions, how="left")
X = df[features]
Y = df[conds_ratios_label]
# need to convert strings to ints because tensors can't mix strings with floats/ints
mapping_1 = {"a": 1, "b": 2}
mapping_2 = {"1": 1, "2": 2}
Y.replace({"condition_1": mapping_1}, inplace=True)
Y.replace({"condition_2": mapping_2}, inplace=True)
X = tf.convert_to_tensor(X)
Y = tf.convert_to_tensor(Y)
model = my_model(input_shape=len(features))
model.fit(X, Y, epochs=1, batch_size=64)
print()
print(model.evaluate(X, Y))
def custom_loss(conditions, true_ratios, y_pred):
y_pred = tf.sigmoid((y_pred - 0.5) * 1000)
uniques, idx, count = gen_array_ops.unique_with_counts_v2(conditions, [0])
num_unique = tf.size(count)
sums = tf.math.unsorted_segment_sum(data=y_pred, segment_ids=idx, num_segments=num_unique)
lengths = tf.cast(count, tf.float32)
pred_ratios = tf.divide(sums, lengths)
mean_pred_ratios = tf.math.reduce_mean(pred_ratios)
mean_true_ratios = tf.math.reduce_mean(true_ratios)
diff = mean_pred_ratios - mean_true_ratios
return K.mean(K.abs(diff))
def standard_loss(y_true, y_pred):
return tf.losses.binary_crossentropy(y_true=y_true, y_pred=y_pred)
def joint_loss(conds_ratios_label, y_pred):
y_true = conds_ratios_label[:, 3]
true_ratios = conds_ratios_label[:, 2]
conditions = tf.gather(conds_ratios_label, [0, 1], axis=1)
loss_1 = standard_loss(y_true=y_true, y_pred=y_pred)
loss_2 = custom_loss(conditions=conditions, true_ratios=true_ratios, y_pred=y_pred)
return 0.5 * loss_1 + 0.5 * loss_2
def my_model(input_shape=None):
model = Sequential()
model.add(Dropout(0, input_shape=(input_shape,)))
model.add(Dense(units=2, activation="relu"))
model.add(Dense(units=1, activation='sigmoid'))
model.add(Flatten())
model.compile(loss=joint_loss, optimizer="Adam",
metrics=[joint_loss, "accuracy"], # had to remove custom_loss because it takes 3 args now
run_eagerly=True)
return model
if __name__ == '__main__':
main()
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主要更新是custom_loss. 我删除了创建 true_ratios DataFramecustom_loss并将其附加到我的Ymain.c 文件中。现在custom_loss需要 3 个参数,其中之一是true_ratios张量。我必须使用gen_array_ops.unique_with_counts_v2和unsorted_segment_sum来获取每组条件的总和。然后我得到了每个组的长度以便创建pred_ratios(根据 计算出每组的比率y_pred)。最后,我得到平均预测比率和平均真实比率,并取绝对差来得到我的自定义损失。
一些注意事项:
tf.round,但我意识到这是不可微分的。所以我把它替换为 的y_pred = tf.sigmoid((y_pred - 0.5) * 1000)内部custom_loss。这本质上是将所有y_pred值取为 0 和 1,但是以可微分的方式。不过,这似乎有点“黑客”,所以如果您对此有任何反馈,请告诉我。run_eagerly=True有效model.compile()。否则我会收到此错误:“ValueError:尺寸必须相等,但...的尺寸为 1 和 2”。我不确定为什么会出现这种情况,但错误源于我使用的行tf.unsorted_segment_sum。unique_with_counts_v2实际上还不存在于tensorflow API中,但它存在于源代码中。我需要它能够按多个条件(而不仅仅是单个条件)进行分组。如果您对此有任何一般性反馈或对上述要点的回应,请随时发表评论。
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