Question

有一个优先级列表，如：

> df  <- data.frame(id = c(1,2,3), Google = c(1,1,0), Yahoo = c(1,1,1), Microsoft = c(0,1,1))
> df
  id Google Yahoo Microsoft
1  1      1     1         0
2  2      1     1         1
3  3      0     1         1

来自二进制数据框，如下所示：

> df  <- data.frame(id = c(1,2,3), Google = c(1,1,0), Yahoo = c(0,0,1), Microsoft = c(0,0,0))
> df
  id Google Yahoo Microsoft
1  1      1     0         0
2  2      1     0         0
3  3      0     1         0

如何生成一个新的数据帧，其中列是相同的，但根据优先级，只有优先级最高的列保留1，另一行在每行中取0？

预期结果的例子：

import tensorflow as tf

def load_graph(frozen_graph_filename):
    # We load the protobuf file from the disk and parse it to retrieve the 
    # unserialized graph_def
    with tf.gfile.GFile(frozen_graph_filename, "rb") as f:
        graph_def = tf.GraphDef()
        graph_def.ParseFromString(f.read())

    # Then, we import the graph_def into a new Graph and returns it 
    with tf.Graph().as_default() as graph:
        # The name var will prefix every op/nodes in your graph
        # Since we load everything in a new graph, this is not needed
        tf.import_graph_def(graph_def, name="prefix")
    return graph

graph = load_graph('tensorflow_inception_graph.pb')
inception_graph = tf.get_default_graph()
x = graph.get_tensor_by_name('prefix/input:0')
y = graph.get_tensor_by_name('prefix/avgpool0/reshape:0')
output_feature = tf.stop_gradient(y) # It's an identity function

# Saving
sess = tf.Session()

builder = tf.saved_model.builder.SavedModelBuilder('./test5')
builder.add_meta_graph_and_variables(sess, ["tag"], signature_def_map= {
        "model": tf.saved_model.signature_def_utils.predict_signature_def(
            inputs= {"x": x},
            outputs= {"finalnode": y})
        })
builder.save()

Answer 1

使用：

mc <- max.col(df[-1], ties.method = 'first')
df[-1] <- 0
df[cbind(1:nrow(df), mc + 1)] <- 1

给出：

> df
  id Google Yahoo Microsoft
1  1      1     0         0
2  2      1     0         0
3  3      0     1         0

如果公司列不在优先级顺序中，您可以使用以下命令进行更改：

priority <- c('Google',"Yahoo",'Microsoft')
df <- df[, c(1, match(priority, names(df)))]

Answer 2

我们也可以使用apply函数：

 df[-1]= t(apply(df[-1], 1, function(x)`[<-`(x,-which.max(x),0)))
 df
  id Google Yahoo Microsoft
1  1      1     0         0
2  2      1     0         0
3  3      0     1         0

根据优先级列表生成新列

2 个答案: