vaccines_by_Beijing=vaccines_by_country[vaccines_by_country['vaccines']=='Sinopharm/Beijing'] vaccines_by_Wuhan=vaccines_by_country[vaccines_by_country['vaccines']=='Sinopharm/Wuhan'] vaccines_by_Sinovac=vaccines_by_country[vaccines_by_country['vaccines']=='Sinovac'] vaccines_by_China_0 = vaccines_by_Beijing.append(vaccines_by_Wuhan) vaccines_by_China = vaccines_by_China_0.append(vaccines_by_Sinovac) 以上的代码能否换个方式实现其功能
时间: 2024-02-03 18:01:43 浏览: 31
可以使用 Pandas 的 loc 方法来实现上述代码的功能,示例代码如下:
```
vaccines_by_China = vaccines_by_country.loc[vaccines_by_country['vaccines'].str.contains('Sinopharm/Beijing|Sinopharm/Wuhan|Sinovac')]
```
这行代码会筛选出疫苗名称包含 "Sinopharm/Beijing"、"Sinopharm/Wuhan"、"Sinovac" 的行,并将其赋值给变量 vaccines_by_China。
相关问题
问题三要求每个工位每天生产的时间不能超过16小时且同种类型疫苗生产全部完成之后才能生产另外类型的疫苗,所以首先计算出每个类型疫苗的生产时间,再通过生产顺序累加得到总天数。经过100随机模拟,得到概率大于90%的天数,再在程序最外层嵌套模拟退火算法,寻得最优解。
这是一个比较复杂的问题,需要进行多步处理。首先,我们需要计算出每个类型疫苗的生产时间。我们可以通过以下代码实现:
```python
# 计算每个类型疫苗的生产时间
vaccine_time = []
for vaccine in vaccines:
time = 0
for station in stations:
if vaccine in station['types']:
time += station['time']
vaccine_time.append(time)
```
其中,`vaccines` 是疫苗种类列表,`stations` 是工位列表,每个工位包含 `types` 和 `time` 两个属性,分别表示该工位可以生产的疫苗种类和生产该疫苗需要的时间。
接下来,我们需要确定疫苗生产的顺序。由于同种类型疫苗需要全部完成之后才能生产另外类型的疫苗,我们可以采用拓扑排序的方法来进行排序。以下是实现代码:
```python
# 拓扑排序确定生产顺序
indegree = [0] * len(vaccines)
graph = [[] for _ in range(len(vaccines))]
for i in range(len(vaccines)):
for j in range(len(vaccines)):
if i != j and vaccines[i] in stations[j]['types']:
indegree[i] += 1
graph[j].append(i)
queue = []
for i in range(len(vaccines)):
if indegree[i] == 0:
queue.append(i)
order = []
while queue:
cur = queue.pop(0)
order.append(cur)
for next_ in graph[cur]:
indegree[next_] -= 1
if indegree[next_] == 0:
queue.append(next_)
```
接下来,我们可以通过累加每个疫苗生产的时间,得到总天数。但是由于每个工位每天生产的时间不能超过16小时,我们需要在累加的过程中进行限制。以下是实现代码:
```python
# 计算总天数
days = 0
cur_time = [0] * len(stations)
for i in order:
time = vaccine_time[i]
while time > 0:
available_stations = []
for j in range(len(stations)):
if vaccines[i] in stations[j]['types'] and cur_time[j] + time <= 16:
available_stations.append(j)
if not available_stations:
cur_time = [0] * len(stations)
days += 1
else:
best_station = available_stations[0]
for j in available_stations:
if stations[j]['time'] < stations[best_station]['time']:
best_station = j
cur_time[best_station] += time
time = 0
days += 1
```
以上代码中,`cur_time` 记录每个工位当前已经使用的时间,如果一个工位的生产时间加上当前已经使用的时间超过了16小时,则需要换一天。在选择可用的工位时,我们优先选择生产时间更短的工位。
最后,我们可以通过随机模拟的方法得到天数大于等于目标天数的概率。以下是实现代码:
```python
# 随机模拟得到概率大于90%的天数
target = 30
count = 0
for _ in range(100):
cur_time = [0] * len(stations)
for i in order:
time = vaccine_time[i]
while time > 0:
available_stations = []
for j in range(len(stations)):
if vaccines[i] in stations[j]['types'] and cur_time[j] + time <= 16:
available_stations.append(j)
if not available_stations:
cur_time = [0] * len(stations)
break
else:
best_station = available_stations[0]
for j in available_stations:
if stations[j]['time'] < stations[best_station]['time']:
best_station = j
cur_time[best_station] += time
time = 0
if sum(cur_time) > 16:
cur_time = [0] * len(stations)
if sum(cur_time) > 0:
count += 1
prob = count / 100
while prob < 0.9:
target += 1
count = 0
for _ in range(100):
cur_time = [0] * len(stations)
for i in order:
time = vaccine_time[i]
while time > 0:
available_stations = []
for j in range(len(stations)):
if vaccines[i] in stations[j]['types'] and cur_time[j] + time <= 16:
available_stations.append(j)
if not available_stations:
cur_time = [0] * len(stations)
break
else:
best_station = available_stations[0]
for j in available_stations:
if stations[j]['time'] < stations[best_station]['time']:
best_station = j
cur_time[best_station] += time
time = 0
if sum(cur_time) > 16:
cur_time = [0] * len(stations)
if sum(cur_time) > 0:
count += 1
prob = count / 100
```
最后,我们可以在程序最外层嵌套模拟退火算法,寻找最优解。但是由于模拟退火算法需要定义评价函数和转移函数,这里就不再展开了。
疫苗预约管理系统基础代码c语言
以下是一个简单的疫苗预约管理系统的基础代码示例,可以作为参考:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_VACCINES 10
#define MAX_APPOINTMENTS 100
struct Appointment {
char name[50];
int age;
char date[20];
char time[20];
};
struct Vaccine {
char name[50];
int quantity;
struct Appointment appointments[MAX_APPOINTMENTS];
int num_appointments;
};
struct VaccineBookingSystem {
struct Vaccine vaccines[MAX_VACCINES];
int num_vaccines;
};
void add_vaccine(struct VaccineBookingSystem* system, char* name, int quantity) {
if (system->num_vaccines == MAX_VACCINES) {
printf("Cannot add more vaccines.\n");
return;
}
struct Vaccine vaccine;
strcpy(vaccine.name, name);
vaccine.quantity = quantity;
vaccine.num_appointments = 0;
system->vaccines[system->num_vaccines] = vaccine;
system->num_vaccines++;
}
int make_appointment(struct VaccineBookingSystem* system, char* vaccine_name, struct Appointment appointment) {
for (int i = 0; i < system->num_vaccines; i++) {
if (strcmp(system->vaccines[i].name, vaccine_name) == 0) {
if (system->vaccines[i].quantity > 0) {
system->vaccines[i].quantity--;
system->vaccines[i].appointments[system->vaccines[i].num_appointments] = appointment;
system->vaccines[i].num_appointments++;
return 1;
} else {
return 0;
}
}
}
return -1;
}
int main() {
struct VaccineBookingSystem system;
system.num_vaccines = 0;
add_vaccine(&system, "Pfizer", 100);
add_vaccine(&system, "Moderna", 50);
struct Appointment appointment;
strcpy(appointment.name, "John");
appointment.age = 35;
strcpy(appointment.date, "2022-05-15");
strcpy(appointment.time, "10:00");
int result = make_appointment(&system, "Pfizer", appointment);
if (result == 1) {
printf("Appointment made successfully.\n");
} else if (result == 0) {
printf("Vaccine is out of stock.\n");
} else {
printf("Vaccine not found.\n");
}
return 0;
}
```
上述代码使用结构体表示预约和疫苗,并使用结构体数组表示整个预约系统中的所有疫苗。在此基础上,实现了两个函数:add_vaccine用于添加新的疫苗,make_appointment用于为指定的疫苗进行预约。
在这个系统中,您可以使用add_vaccine函数添加新的疫苗,并使用make_appointment函数为指定的疫苗进行预约。如果预约成功,该预约将被添加到疫苗的预约列表中,并且该疫苗的数量将减少1。如果预约失败(例如因为该疫苗已经用尽),则返回0。如果指定的疫苗不存在,则返回-1。
请注意,这只是一个基础的示例代码,实际的疫苗预约管理系统需要考虑更多的因素,例如安全性、预约时间窗口、用户身份验证等。
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