如何更新给定索引的行/列单元格值？

Question

如果有重复项，我试图覆盖1的列，但是由于某种原因，此代码不会覆盖1的列，它会执行其他操作。

样本输入

dhid,midx,midy,midz,a,dtype

AAA1,321.235,200.436,59.72,7,RR

AAA7,321.235,200.436,59.72,-99,CR

AAA2,321.235,200.236,68.7,15,CR

示例输出：

dhid,midx,midy,midz,a,dtype,KEEPVA

AAA1,321.235,200.436,59.72,7,RR,1

AAA7,321.235,200.436,59.72,-99,CR,0  --> GETS RID OF THIS DUPLICATE THAT IS MISSING

AAA2,321.235,200.236,68.7,15,CR,1

到目前为止我的尝试

    dup_dict = []
    flg = False
    for i in range(npts-1):
        d2 = 0.0
        for idir in range(2,-2,-1):
            if (idir==-1 and d2 <= d2tol ):
                dup_dict.append([i,i+1,d2])
                break
            d = np.abs(xyz[i,idir]-xyz[i+1,idir])
            if (d > dtol):
                break
            d2 = d2 + d ** 2
    print('Total Duplicates Found: %d'%(len(dup_dict)))

    df['KEEPVA'] = np.ones(len(df.index))
    fmtlist = ['%s','%0.5f','%0.5f','%0.5f','%0.5f','%d','%s']
    fl = open('dup_gq.log','w')
    fl.write(','.join(['dhid','midx','midy','midz','a','KEEPVA','dtype'])+'\n')

    for i0,i1,d in dup_dict:
        aufa0 = df['a'].values[i0]
        aufa1 = df['a'].values[i1]
        dtype0 = df['dtype'].values[i0]
        dtype1 = df['dtype'].values[i1]
        if (dtype0!=dtype1):
            print(df['dhid'].values[i0],df['dhid'].values[i1],dtype0,dtype1,a0,a1)
        if a0 > a1:
            df.loc[i1,'KEEPVA'] = 0
        if  a0 <= a1:
            df.loc[i0,'KEEPVA'] = 0

        print(df['dhid'].values[i0],df['dhid'].values[i1],df['KEEPVA'].values[i0],
        df['KEEPVA'].values[i1])    
        a = df[['dhid','midx','midy','midz','a','KEEPVA','dtype']].values[[i0,i1],:]
        np.savetxt(fl,a,fmt=fmtlist,delimiter=',')
    fl.close()