我正在尝试使用 3 格式的电源标记数据点,即24^3
。
例如,
A = [1,2,3,4,5]
B = [100,200,300,400,500]
fig = plt.figure()
ax = fig.add_subplot(111)
plt.plot(A,B)
n=[24,48,96,192]
for i, txt in enumerate(n):
ax.annotate(txt, (A[i],B[i]))
plt.show()
所以,我希望n=[24,48,96,192]
代替24^3,48^3,96^3
,而不是import play.api.libs.json.{JsValue, Json, Writes}
sealed trait Model
case class User(name: String, age: String, address: String) extends Model
object User {
implicit val userWrites = Json.writes[User]
}
case class CallLog(timestamp: String, status_of_call: String) extends Model
object CallLog {
implicit val callLogWrites = Json.writes[CallLog]
}
implicit val modelWrites = new Writes[Model] {
override def writes(o: Model): JsValue = o match {
case u: User => Json.toJson(u)
case cl: CallLog => Json.toJson(cl)
}
}
def helper(models: Model*): Either[JsValue, String] = models match {
case Nil => Right("Error")
case _ => Left(Json.toJson(models))
}
helper(User("John", "32", "..."))
helper(User("John", "32", "..."), CallLog("now", "In progress"))
,如果我可以使用改进的表示形式来表示3的力量,那么我会更好。{/ p>
我怎么能这样做?
答案 0 :(得分:1)
请参阅:http://matplotlib.org/users/mathtext.html
示例:
import matplotlib.pylab as plt
A = [1,2,3,4,5]
B = [100,200,300,400,500]
fig = plt.figure()
ax = fig.add_subplot(111)
plt.plot(A,B)
n=[24,48,96,192]
for i, txt in enumerate(n):
ax.annotate(r'${}^3$'.format(txt), (A[i],B[i]))
plt.show()
答案 1 :(得分:0)
您的注释通话只需要进行一些小改动:
for i, txt in enumerate(n):
ax.annotate(str(txt) + '$^3$', (A[i],B[i]))
使用'$ ... $',您可以输入某种数学模式。 在此处查看更多详细信息:http://matplotlib.org/users/mathtext.html