谷歌Foobar解决方案可以在Jupyter笔记本上运行，但不能在谷歌的终端上运行

Question

我在GoogleFoobar的第3级，我编写的代码在Jupyter笔记本中运行，但是当我在Foobar命令行中运行它时，没有一个测试用例通过。当我在Foobar中运行它时，没有任何错误，它只是说答案不正确

Doomsday Fuel
=============

Making fuel for the LAMBCHOP's reactor core is a tricky process because of the exotic matter involved. It starts as raw ore, then during processing, begins randomly changing between forms, eventually reaching a stable form. There may be multiple stable forms that a sample could ultimately reach, not all of which are useful as fuel. 

Commander Lambda has tasked you to help the scientists increase fuel creation efficiency by predicting the end state of a given ore sample. You have carefully studied the different structures that the ore can take and which transitions it undergoes. It appears that, while random, the probability of each structure transforming is fixed. That is, each time the ore is in 1 state, it has the same probabilities of entering the next state (which might be the same state).  You have recorded the observed transitions in a matrix. The others in the lab have hypothesized more exotic forms that the ore can become, but you haven't seen all of them.

Write a function solution(m) that takes an array of array of nonnegative ints representing how many times that state has gone to the next state and return an array of ints for each terminal state giving the exact probabilities of each terminal state, represented as the numerator for each state, then the denominator for all of them at the end and in simplest form. The matrix is at most 10 by 10. It is guaranteed that no matter which state the ore is in, there is a path from that state to a terminal state. That is, the processing will always eventually end in a stable state. The ore starts in state 0. The denominator will fit within a signed 32-bit integer during the calculation, as long as the fraction is simplified regularly. 

For example, consider the matrix m:
[
  [0,1,0,0,0,1],  # s0, the initial state, goes to s1 and s5 with equal probability
  [4,0,0,3,2,0],  # s1 can become s0, s3, or s4, but with different probabilities
  [0,0,0,0,0,0],  # s2 is terminal, and unreachable (never observed in practice)
  [0,0,0,0,0,0],  # s3 is terminal
  [0,0,0,0,0,0],  # s4 is terminal
  [0,0,0,0,0,0],  # s5 is terminal
]
So, we can consider different paths to terminal states, such as:
s0 -> s1 -> s3
s0 -> s1 -> s0 -> s1 -> s0 -> s1 -> s4
s0 -> s1 -> s0 -> s5
Tracing the probabilities of each, we find that
s2 has probability 0
s3 has probability 3/14
s4 has probability 1/7
s5 has probability 9/14
So, putting that together, and making a common denominator, gives an answer in the form of
[s2.numerator, s3.numerator, s4.numerator, s5.numerator, denominator] which is
[0, 3, 2, 9, 14].

Languages
=========

To provide a Java solution, edit Solution.java
To provide a Python solution, edit solution.py

Test cases
==========
Your code should pass the following test cases.
Note that it may also be run against hidden test cases not shown here.

-- Java cases --
Input:
Solution.solution({{0, 2, 1, 0, 0}, {0, 0, 0, 3, 4}, {0, 0, 0, 0, 0}, {0, 0, 0, 0,0}, {0, 0, 0, 0, 0}})
Output:
    [7, 6, 8, 21]

Input:
Solution.solution({{0, 1, 0, 0, 0, 1}, {4, 0, 0, 3, 2, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}})
Output:
    [0, 3, 2, 9, 14]

-- Python cases --
Input:
solution.solution([[0, 2, 1, 0, 0], [0, 0, 0, 3, 4], [0, 0, 0, 0, 0], [0, 0, 0, 0,0], [0, 0, 0, 0, 0]])
Output:
    [7, 6, 8, 21]

Input:
solution.solution([[0, 1, 0, 0, 0, 1], [4, 0, 0, 3, 2, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]])
Output:
    [0, 3, 2, 9, 14]

我的解决办法如下：

import numpy as np
import pandas as pd
from fractions import Fraction

def solution(m):
    if m == [[0]]:
        return [1, 1]
    else:
        return run_matrix_computation(m)

def run_matrix_computation(starting_ore_matrix):
    numpy_matrix = convert_to_numpy_matrix(starting_ore_matrix)
    ordered_matrix = order_matrix(numpy_matrix)
    absorption_matrix, split_index = create_absorption_matrix(ordered_matrix)
    R, Q = store_R_and_Q(absorption_matrix, split_index)
    FR = compute_FR(R, Q)
    first_row = get_first_row_of_FR(FR)
    common_denominator, fraction_list = calculate_common_denominator(first_row)
    return return_int_array(common_denominator, fraction_list)

def convert_to_numpy_matrix(original_matrix):
    return np.asarray(original_matrix)

def order_matrix(numpy_matrix):
    labeled_dataframe = pd.DataFrame(data=numpy_matrix)
    index_order = labeled_dataframe.sum(axis=1).sort_values(ascending=True).index
    converted_matrix = convert_matrix_to_fractions(labeled_dataframe)
    return converted_matrix.iloc[index_order, index_order]

def convert_matrix_to_fractions(original_matrix):
    for i in range(len(original_matrix.index)):
        sum = original_matrix.sum(axis=1).iloc[i]
        if sum != 0:
            for j in range(len(original_matrix.columns)):
                if original_matrix.iloc[i, j]:
                    original_matrix.iloc[i, j] = original_matrix.iloc[i, j] / sum
    return original_matrix

def create_absorption_matrix(sorted_matrix):
    count = 0
    for i in range(len(sorted_matrix.index)):
        if not(sorted_matrix.sum(axis=1).iloc[i]):
            sorted_matrix.iloc[i, i] = 1
            count = count + 1
    return sorted_matrix, count

def store_R_and_Q(absorption_matrix, split_index):
    return split_into_new_matrices(absorption_matrix, split_index)

def split_into_new_matrices(absorption_matrix, split_index):
    numpy_matrix = absorption_matrix.to_numpy()
    R = numpy_matrix[split_index:, :split_index]
    Q = numpy_matrix[split_index:, split_index:]
    return R, Q

def calculate_F(R, Q):
    num_rows, num_cols = Q.shape
    I = np.identity(num_rows)
    IQ = I - Q
    return np.linalg.inv(IQ)

def compute_FR(R, Q):
    F = calculate_F(R, Q)
    return np.matmul(F, R)

def get_first_row_of_FR(FR):
    return FR[0, :]

def calculate_common_denominator(list):
    fraction_list = convert_to_fractions(list)
    list_denominators = get_denominators(fraction_list)
    GCD = calculate_greatest_common_denominator(list_denominators)
    return GCD, fraction_list

def convert_to_fractions(list):
    fraction_list = []
    for i in range(len(list)):
        fraction_list.append(Fraction(list[i]).limit_denominator())
    return fraction_list

def get_denominators(fractions):
    denominators = []
    for i in range(len(fractions)):
        denominators.append(fractions[i].denominator)
    return denominators

def calculate_greatest_common_denominator(denominators):
    GCD = 0
    if len(denominators) == 1:
        return denominators
    else:
        for i in range(len(denominators) - 1):
            cur_GCD = np.lcm(denominators[i], denominators[i + 1])
            if cur_GCD > GCD:
                GCD = cur_GCD
    return GCD

def return_int_array(denominator, fractions):
    final_list = []
    for i in range(len(fractions)):
        if(not fractions[i].numerator):
            final_list.append(0)
        else:
            multiplier = denominator/fractions[i].denominator
            final_list.append(int(fractions[i].numerator * multiplier))
    final_list.append(int(denominator))
    return final_list

用这段代码运行任何测试用例都是可行的，但是每个测试都会在Foobar上失败。是否存在某种格式错误？我已经检查了返回的对象类型和Foobar正在查找的对象类型，它们都是int列表。据我所知，我的代码中的所有内容都得到了Python 2.7.13的支持，而这正是Foobar所使用的。我使用的图书馆也是允许的

Answer 1

我已经晚了，但我相信问题在于Python2.7如何处理除法。我把它放在我的文件的开头，这样就解决了这个问题

from __future__ import division

Answer 2

我认为您应该首先检查库是否已安装以及版本。 然后您应该使用replit或其他与确切环境相关的东西。 最后，您的代码是否在解决方案类中