如何在 tensorflow 中将字典转换为张量 [英] How to convert a dictionary into a tensor in tensorflow

问题描述

这是我的字典:

docs = {'computer': {'1': 1, '3': 5, '8': 2},
        'politics': {'0': 2, '1': 2, '3': 1}}

我想像这样创建一个 9 * 2 张量:

I want to create a 9 * 2 tensor like this:

[
    [0, 1, 0, 5, 0, 0, 0, 0, 2],
    [2, 2, 0, 1, 0, 0, 0, 0, 0, 0]
]

这里，因为最大项目是 8，所以我们有 9 行.但是，可以根据字典增加行数和列数.

Here, because the max item is 8 so we have 9 rows. But, the number of rows and columns can increase based on the dictionary.

我已经尝试使用 for-loop 来实现这一点，但由于字典很大，它根本没有效率，而且它使用我需要的列表来实现 tensor.

I have tried to implement this using for-loop though as the dictionary is big it's not efficient at all and also it implemented using the list I need that to be a tensor.

maxr = 0
for i, val in docs.items():
    for j in val.keys():
        if int(j) > int(maxr):
            maxr = int(j)

final_lst = []
for val in docs.values():
    lst = [0] * (maxr+1)
    for j, val2 in sorted(val.items()):
        lst[int(j)] = val2
    final_lst.append(lst)

print(final_lst)

推荐答案

如果您可以使用 pandas 和 numpy，这里是您可以使用的方法.

If you are ok with using pandas and numpy, here's how you can do it.

import pandas as pd 
import numpy as np 

# Creates a dataframe with keys as index and values as cell values. 
df  = pd.DataFrame(docs)

# Create a new set of index from min and max of the dictionary keys. 
new_index = np.arange( int(df.index.min()),
                       int(df.index.max())).astype(str)

# Add the new index to the existing index and fill the nan values with 0, take a transpose of dataframe. 

new_df = df.reindex(new_index).fillna(0).T.astype(int)

#          0  1  2  3  4  5  6  7
#computer  0  1  0  5  0  0  0  0
#politics  2  2  0  1  0  0  0  0

如果你只想要数组，你可以调用array = new_df.values.

If you just want the array, you can call array = new_df.values.

#[[0 1 0 5 0 0 0 0]
# [2 2 0 1 0 0 0 0]]

如果你想要张量，那么你可以使用 tf.convert_to_tensor(new_df.values)

If you want tensor, then you can use tf.convert_to_tensor(new_df.values)

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