편집 기록

편집 기록
  • 프로필 nowp님의 편집
    날짜2023.04.01

    파이썬 코드 질문


    # error for DatN[0]
    datX = DatN[0][0]   # -13.6
    datY = DatN[0][1]   # -22.8
    error = (a*datX - datY)**2     # 84.2
    print("SSE; error for DatN[0]=",round(error,1))
    
    #########################
    
    
    
    # SSE
    # sum of errors for all data
    SSE=0
    for i in range (0,Num):
       datX = DatN[i][0]   
       datY = DatN[i][1]   
       error = (a*datX - datY)**2    
       SSE += error
       print(i,": error=",round(error,1),"  SSE=",round(SSE,1))
       input()
    #for.i
    print("Out of for.i;","  SSE=",round(SSE,1))    #1570.5
    input()
    

    함수 만들기; Compute_SSE()

    # ============  Compute_SSE()  =============
    def Compute_SSE(DatN):
      # SSE
      # sum of errors for all data
      SSE=0
      for i in range (0,Num):
        datX = DatN[i][0]   
        datY = DatN[i][1]   
        error = (a*datX - datY)**2    
        SSE += error
        #print(i,": error=",round(error,1),"  SSE=",round(SSE,1))
        #input()
      #for.i
      #print("Out of for.i;","  SSE=",round(SSE,1))    #1570.5
      #input()
      return SSE
    #compute_sse()
    

    솔루션을 찾는 후속코드 만들어주시면 감사하겠습니다.

  • 프로필 엽토군님의 편집
    날짜2023.03.30

    파이썬 코드 질문


        # error for DatN[0]
    datX = DatN[0][0]   # -13.6
    datY = DatN[0][1]   # -22.8
    error = (a*datX - datY)**2     # 84.2
    print("SSE; error for DatN[0]=",round(error,1))
    
    #########################
    
    
    
        # SSE
         # sum of errors for all data
    SSE=0
    for i in range (0,Num):
       datX = DatN[i][0]   
       datY = DatN[i][1]   
       error = (a*datX - datY)**2    
       SSE += error
       print(i,": error=",round(error,1),"  SSE=",round(SSE,1))
       input()
    #for.i
    print("Out of for.i;","  SSE=",round(SSE,1))    #1570.5
    input()
    
    
    함수 만들기; Compute_SSE()
    
    #============  Compute_SSE()  =============
    def Compute_SSE(DatN):
        # SSE
         # sum of errors for all data
      SSE=0
      for i in range (0,Num):
        datX = DatN[i][0]   
        datY = DatN[i][1]   
        error = (a*datX - datY)**2    
        SSE += error
        #print(i,": error=",round(error,1),"  SSE=",round(SSE,1))
        #input()
      #for.i
      #print("Out of for.i;","  SSE=",round(SSE,1))    #1570.5
      #input()
      return SSE
    #compute_sse()
    

    솔루션을 찾는 후속코드 만들어주시면 감사하겠습니다.

  • 프로필 wlgns528@gmail.com님의 편집
    날짜2023.03.29

    파이썬 코드 질문


    # error for DatN[0]
    

    datX = DatN[0][0] # -13.6 datY = DatN[0][1] # -22.8 error = (a*datX - datY)**2 # 84.2 print("SSE; error for DatN[0]=",round(error,1))

    # SSE
     # sum of errors for all data
    

    SSE=0 for i in range (0,Num): datX = DatN[i][0]
    datY = DatN[i][1]
    error = (a*datX - datY)**2
    SSE += error print(i,": error=",round(error,1)," SSE=",round(SSE,1)) input()

    for.i

    print("Out of for.i;"," SSE=",round(SSE,1)) #1570.5 input()

    함수 만들기; Compute_SSE()

    ============ Compute_SSE() =============

    def Compute_SSE(DatN): # SSE # sum of errors for all data SSE=0 for i in range (0,Num): datX = DatN[i][0]
    datY = DatN[i][1]
    error = (a*datX - datY)**2
    SSE += error #print(i,": error=",round(error,1)," SSE=",round(SSE,1)) #input() #for.i #print("Out of for.i;"," SSE=",round(SSE,1)) #1570.5 #input() return SSE

    compute_sse()

    솔루션을 찾는 후속코드 만들어주시면 감사하겠습니다.