python - 如何修复 'TypeError: cannot convert dictionary update sequence element #0 to a sequence'

Question

我正在查看 Google 的免费机器学习速成类(class)，并尝试根据他们类(class)的第一部分制作一个预测模型。但是，在输入函数中，有一个字典，我不断收到此错误，

in my_input_fn
    features = {key:np.array(value) for key,value in dict(features).items()}
ValueError: dictionary update sequence element #0 has length 59; 2 is required

我尝试 reshape 和修改我的“targets”和“my_features”变量，它确实解决了第一个错误，但现在我收到此错误，

TypeError: cannot convert dictionary update sequence element #0 to a sequence

import numpy as np
import pandas as pd
import tensorflow as tf
from tensorflow.python.data import Dataset

tf.logging.set_verbosity(tf.logging.ERROR)
pd.options.display.max_rows = 10
pd.options.display.float_format = '{:.1f}'.format

world_gdp_dataset = pd.read_csv("/Users/usr/Desktop/file/API_NY/API_NY.GDP.MKTP.CD_DS2_en_csv_v2_40924.csv", sep=",", skiprows=2, names=["Country Name","Country Code","Indicator Name","Indicator Code","1960","1961","1962","1963","1964","1965","1966","1967","1968","1969","1970","1971","1972","1973","1974","1975","1976","1977","1978","1979","1980","1981","1982","1983","1984","1985","1986","1987","1988","1989","1990","1991","1992","1993","1994","1995","1996","1997","1998","1999","2000","2001","2002","2003","2004","2005","2006","2007","2008","2009","2010","2011","2012","2013","2014","2015","2016","2017","2018"])

my_data = world_gdp_dataset.iloc[[29], 4:]

br_columns = []
for num in world_gdp_dataset.iloc[29]:
    br_columns.append(num)
my_features = br_columns[4:]

targets = world_gdp_dataset.columns[4:]
targets = [int(ind) for ind in targets]

my_optimizer = tf.train.GradientDescentOptimizer(learning_rate=0.0000001)
my_optimizer = tf.contrib.estimator.clip_gradients_by_norm(my_optimizer, 5.0)

linear_regressor = tf.estimator.LinearRegressor(feature_columns = my_features, optimizer = my_optimizer)

def my_input_fn(features, targets, batch_size=1, shuffle=True, num_epochs=None):
    features = {key:np.array(value) for key,value in dict(features).items()}

    ds = Dataset.from_tensor_slices((features, targets))#WARNING: 2GB limit
    ds = ds.batch(batch_size).repeat(num_epochs)

    if shuffle:
        ds = ds.shuffle(buffer_size=10000)

    features, labels = ds.make_one_shot_iterator().get_next()
    return features, labels

_ = linear_regressor.train(input_fn = lambda:my_input_fn(my_features, targets), steps = 100)

如果我要打印目标，它会返回，

[1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018]

当我打印 my_features 时它会返回，

[[1.51655699e+10 1.52368549e+10 1.99262938e+10 2.30214773e+10
  2.12118923e+10 2.17900351e+10 2.70627166e+10 3.05918341e+10
  3.38758819e+10 3.74588982e+10 4.23276001e+10 4.92044567e+10
  5.85390088e+10 7.92790577e+10 1.05136008e+11 1.23709377e+11
  1.52678020e+11 1.76171284e+11 2.00800892e+11 2.24969489e+11
  2.35024599e+11 2.63561089e+11 2.81682304e+11 2.03304515e+11
  2.09023913e+11 2.22942790e+11 2.68137225e+11 2.94084112e+11
  3.30397382e+11 4.25595310e+11 4.61951782e+11 6.02860000e+11
  4.00599250e+11 4.37798578e+11 5.58111997e+11 7.69305386e+11
  8.50426433e+11 8.83199625e+11 8.63723412e+11 5.99388580e+11
  6.55420645e+11 5.59372276e+11 5.07962488e+11 5.58319921e+11
  6.69316654e+11 8.91630177e+11 1.10764029e+12 1.39708435e+12
  1.69582457e+12 1.66701978e+12 2.20887165e+12 2.61620158e+12
  2.46518867e+12 2.47280646e+12 2.45599405e+12 1.80221437e+12
  1.79627544e+12 2.05359497e+12 1.86862609e+12]]

Answer 1

你想在这行做什么？

看起来好像您正在尝试从现有字典构建新字典，但您正在将值转换为 numpy 数组。

你能不能说:

features = dict(features)
for key in features.keys():
    features[key] = np.array(features[key])

如果这不是您想要做的，请进一步扩展。

我认为您可能正在尝试按照以下方式做一些事情:

new_dict = dict(features)
[features.update({key, np.array(val)} for key, val in new_dict.items()]

这使用列表理解。参见这里:https://www.pythonforbeginners.com/basics/list-comprehensions-in-python