我需要创建一个函数，当给定边长时，该函数将生成菱形。菱形应包含0和1的数组
到目前为止，我已经弄清楚了如何制作菱形，但我不知道如何为不同的边长编程功能
到目前为止，我有： 边长为3
public class MessageBuilder {
private final EventPayload messagePayload;
public MessageBuilder(){
messagePayload = new EventPayload();
}
public EventPayload build(Activity event) {
createBaseActivity(event);
if (event instanceof Member) {
updateToMemberActivity(event);
return messagePayload;
}
if (event instanceof MeetingActivity) {
updateToMeetingActivity(event);
return messagePayload;
}
if (event instanceof ServiceActivity) {
updateToServiceActivity(event);
return messagePayload;
}
if (event instanceof LetterActivity) {
updateToLetterActivity(event);
return messagePayload;
}
if (event instanceof ReceiptActivity) {
updateToReceiptActivity(event);
return messagePayload;
}
return messagePayload;
}
private EventPayload createBaseActivity(Activity event) {
messagePayload.setId(event.getId());
messagePayload.setMemberId(event.getMemberId());
messagePayload.setActivityDateTime(event.getActivityDateTime());
messagePayload.setMemberIdIsPrimaryId(event.isMemberIdIsPrimaryId());
messagePayload.setBuCode(event.getBuCode());
messagePayload.setStateCode(event.getStateCode());
messagePayload.setProductCode(event.getProductCode());
messagePayload.setBusinessUnit(event.getBusinessUnit());
messagePayload.setActivityType(event.getActivityType());
messagePayload.setActivityIntent(event.getActivityIntent());
messagePayload.setActivityAction(event.getActivityAction());
messagePayload.setActivityStatus(event.getActivityStatus());
messagePayload.setSourceSystem(event.getSourceSystem());
messagePayload.setCaseId(event.getCaseId());
messagePayload.setInitiatedBy(event.getInitiatedBy());
messagePayload.setActivityDetailsPointer(event.getActivityDetailsPointer());
messagePayload.setBusinessLine(event.getBusinessLine());
messagePayload.setEventInstanceType(EventInstanceType.ACTIVITY.name());
return messagePayload;
}
private EventPayload updateToMemberActivity(Activity event) {
Member subTypedEvent = (Member) event;
messagePayload.setDeliveryType(subTypedEvent.getDeliveryType());
messagePayload.setDocumentType(subTypedEvent.getDocumentType());
messagePayload.setEventInstanceType(EventInstanceType.MEMBER.name());
return messagePayload;
}
private EventPayload updateToMeetingActivity(Activity event) {
MeetingActivity subTypedEvent = (MeetingActivity) event;
messagePayload.setDeliveryType(subTypedEvent.getDeliveryType());
messagePayload.setDocumentType(subTypedEvent.getDocumentType());
messagePayload.setEventInstanceType(EventInstanceType.MEETING.name());
return messagePayload;
}
private EventPayload updateToServiceActivity(Activity event) {
ServiceActivity subTypedEvent = (ServiceActivity) event;
messagePayload.setReceivedBy(subTypedEvent.getReceivedBy());
messagePayload.setConcernedParty(subTypedEvent.getConcernedParty());
messagePayload.setSummary(subTypedEvent.getSummary());
messagePayload.setEventInstanceType(EventInstanceType.SERVICE.name());
return messagePayload;
}
private EventPayload updateToLetterActivity(Activity event) {
LetterActivity subTypedEvent = (LetterActivity) event;
messagePayload.setDeliveryType(subTypedEvent.getDeliveryType());
messagePayload.setDocumentType(subTypedEvent.getDocumentType());
messagePayload.setPaymentType(subTypedEvent.getPaymentType());
messagePayload.setEventInstanceType(EventInstanceType.LETTER.name());
return messagePayload;
}
private EventPayload updateToReceiptActivity(Activity event) {
ReceiptActivity subTypedEvent = (ReceiptActivity) event;
messagePayload.setPaymentType(subTypedEvent.getPaymentType());
messagePayload.setEventInstanceType(EventInstanceType.RECEIPT.name());
return messagePayload;
}
}
如何编写不同长度的函数
同样，如果给定的长度为1，它应该返回import numpy as np
#line1
a=np.zeros(3+2)
a[3-1]=1
#line2
b=np.zeros(3+2)
b[3-2]=1
b[3]=1
#line3
c=np.zeros(3+2)
c[3-3]=1
c[3+1]=1
print(np.concatenate((a,b,c,b,a),axis=1).reshape(5,5))
任何反馈将不胜感激
更新：我认为循环可能会找出行数
答案 0 :(得分：1)
您可以使用水平和垂直图案的交集来做到这一点：
import numpy as np
N = 5
H = abs(np.arange(1-N,N+1,2))//2
V = (N-1)//2-H[:,None]
diamond = (H==V)*1
print(diamond)
[[0 0 1 0 0]
[0 1 0 1 0]
[1 0 0 0 1]
[0 1 0 1 0]
[0 0 1 0 0]]
在视觉上，这对应于行和列之间相交的数字相等：
对于N = 7：
[3, 2, 1, 0, 1, 2, 3]
0 . . . x . . .
1 . . x . x . .
2 . x . . . x .
3 x . . . . . x
2 . x . . . x .
1 . . x . x . .
0 . . . x . . .
对于N = 8：
[3, 2, 1, 0, 0, 1, 2, 3]
0 . . . x x . . .
1 . . x . . x . .
2 . x . . . . x .
3 x . . . . . . x
3 x . . . . . . x
2 . x . . . . x .
1 . . x . . x . .
0 . . . x x . . .
如果要填充钻石，请使用diamond = (H<=V)*1
答案 1 :(得分：0)
我花了更长的时间，所以我可以扩展功能以处理其他几何形状
import numpy as np
def diamondarray(dimension=1):
#// initialize 2d array
a=np.zeros((dimension,dimension))
#// find the middle of the array
midpoint=(dimension-1)/2
#// initialize an offset
offset=-1
offsetstep=1
#// loop through rows and columns
for row in range(dimension):
if dimension%2 == 0 and row == np.ceil(midpoint):
#// repeat offset for second midpoint row
offset=offset
else:
if row <= np.ceil(midpoint):
#// increase offset for each row for top
offset=offset+offsetstep
else:
#// decrease offset for each row for bottom
offset=offset-offsetstep
for col in range(dimension):
#// set value to one
if dimension%2 == 0:
if col <= np.floor(midpoint):
if col == np.floor(midpoint)-offset:
a[row,col]=fill
if col >= np.ceil(midpoint):
if col == int(midpoint)+offset+1:
a[row,col]=fill
else:
if col == midpoint+offset or col == midpoint-offset:
pass
a[row,col]=fill
return a
对于N = 5：
print（diamondarray（5））
[[0. 0. 1. 0. 0.]
[0. 1. 0. 1. 0.]
[1. 0. 0. 0. 1.]
[0. 1. 0. 1. 0.]
[0. 0. 1. 0. 0.]]
对于N = 8：
print（diamondarray（8））
[[0. 0. 0. 1. 1. 0. 0. 0.]
[0. 0. 1. 0. 0. 1. 0. 0.]
[0. 1. 0. 0. 0. 0. 1. 0.]
[1. 0. 0. 0. 0. 0. 0. 1.]
[1. 0. 0. 0. 0. 0. 0. 1.]
[0. 1. 0. 0. 0. 0. 1. 0.]
[0. 0. 1. 0. 0. 1. 0. 0.]
[0. 0. 0. 1. 1. 0. 0. 0.]]