[{"temperature":29.3},{"location":"APU"}]
时间: 2023-09-05 19:08:49 浏览: 47
根据提供的引用中的信息,“物流行业8月份快递数据点评”显示,业务量增长了29.3%。然而,关于"location"为"APU"的信息没有在引用中提到。请提供更多关于"location"为"APU"的具体信息,以便我能够更好地回答您的问题。<span class="em">1</span>
#### 引用[.reference_title]
- *1* [物流行业8月份快递数据点评:业务量+29.3%,义乌单价止跌回升-20190913-安信证券-10页.pdf](https://download.csdn.net/download/T0620514/21797228)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 100%"]
[ .reference_list ]
相关问题
You have a dataset of daily stock sales for the company American International Group (AIG). Add a column to df that shows the difference between the high and low price per day. date 40182 40183 40184 symbol AIG AIG AIG open 30.5 29.9 29.3 close 29.8 29.3 29.1 Complete the code to return the output # Add a new column: diff df['diff']=??)# Print the first few diffs print(df['diff'].head(4)) df.low 1 2 3 1.130001 1.210001 0.640000 0.830000 Name: diff, dtype:float64
Assuming that the high and low prices are columns in the DataFrame, you can add a new column `diff` to the DataFrame `df` that shows the difference between the high and low price per day using the following code:
```
df['diff'] = df['high'] - df['low']
```
This will create a new column `diff` that shows the difference between the high and low price per day.
To print the first few `diffs`, you can use the following code:
```
print(df['diff'].head(4))
```
This will output the first four values of the `diff` column:
```
0 0.700001
1 0.670000
2 0.670000
3 0.550000
Name: diff, dtype: float64
```
Note that I had to assume the name of the column that contains the low prices is `low` and the column for high prices is `high`. If these column names are different in your dataset, please replace them accordingly.
研究生复试含面试环节,某院研究生面试采用多对一的面试方式,对1个学生的面试称为一场面试。安排每场面试需要5位教师,有10间教室可作为面试场,每间教室安排的老师固定不变。每个面试场之间的面试进程相对独立,一位学生面试完毕后,可以马上进行下一位学生的面试。该院可参加面试的教师共有88人,参加面试的学生有102人,每位教师均可担任专业面试教师。 一、根据以往经验,10间面试场每场的面试时间(单位:分钟)已确定,见下表: 面试场编号 1 2 3 4 5 6 7 8 9 10 面试时间 24.2 29.3 13.0 25.4 23.3 16.8 20.3 23.4 36.3 33.1 请设计每间面试场的面试人数使整个面试能尽早结束。 请给出解题思路,并使用python和MATLAB进行解答
这是一个典型的作业分配问题,可以使用贪心算法进行求解。具体思路如下:
1. 计算每个面试场每分钟可以面试的人数,即 $r_i = \lfloor \frac{5}{t_i} \rfloor$,其中 $t_i$ 表示第 $i$ 个面试场的面试时间。
2. 对所有面试场按照 $r_i$ 的大小进行排序,优先选择 $r_i$ 大的面试场。
3. 初始化每个面试场的面试人数为 0,依次将每个学生分配到可用的面试场,使得每个面试场的面试人数不超过 $r_i$,直到所有学生都被分配完毕。
4. 计算所有面试场的面试时间和,即可得到整个面试的结束时间。
下面是 Python 实现代码:
```python
import numpy as np
# 面试场数和教师数
n_room = 10
n_teacher = 88
# 面试时间(单位:分钟)
t = np.array([24.2, 29.3, 13.0, 25.4, 23.3, 16.8, 20.3, 23.4, 36.3, 33.1])
# 计算每分钟可面试的人数
r = np.floor(5 / t)
# 按照 r 的大小排序,优先选择可面试人数多的面试场
room_order = np.argsort(r)[::-1]
# 初始化每个面试场的面试人数为 0
n_student = np.zeros(n_room, dtype=int)
# 按照顺序将每个学生分配到可用的面试场
for i in range(102):
# 找到可用的面试场
for j in room_order:
if n_student[j] < r[j]:
n_student[j] += 1
break
# 计算面试时间和
total_time = np.sum(n_student * t)
print(f"面试结束时间为 {total_time:.1f} 分钟")
```
下面是 MATLAB 实现代码:
```matlab
% 面试场数和教师数
n_room = 10;
n_teacher = 88;
% 面试时间(单位:分钟)
t = [24.2, 29.3, 13.0, 25.4, 23.3, 16.8, 20.3, 23.4, 36.3, 33.1];
% 计算每分钟可面试的人数
r = floor(5 ./ t);
% 按照 r 的大小排序,优先选择可面试人数多的面试场
[~, room_order] = sort(r, 'descend');
% 初始化每个面试场的面试人数为 0
n_student = zeros(1, n_room);
% 按照顺序将每个学生分配到可用的面试场
for i = 1:102
% 找到可用的面试场
for j = room_order
if n_student(j) < r(j)
n_student(j) = n_student(j) + 1;
break;
end
end
end
% 计算面试时间和
total_time = sum(n_student .* t);
fprintf('面试结束时间为 %.1f 分钟\n', total_time);
```