Question

我需要将下面的数组分成三个数组： tc_excel变量的结果

[(1000000, ['FA'], 'bev.xml'),
 (1001001, ['TC', 'CT03', 'CT04'], False),
 (1003000, ['FA'], 'phev.xml'),
 (1003001, ['TC', 'CT01', 'CT03', 'CT04'], False),
 (1003002, ['TC', 'CT01', 'CT03', 'CT04'], False),
 (1004000, ['FA'], 'tesla.xml'),
 (1004001, ['TC', 'CT03', 'CT04'], False),
 (1004002, ['TC', 'CT03', 'CT04'], False)]

一个数组应该是所有在文件名中带有“ bev”的xml测试用例。
第二个数组应在xml文件名中具有phev
其余应该是所有其他情况

所有拆分后的数组应至少有一个['FA']其中的第一个，（ ['1000000,'1003000,'1004000]

def testcasefilter(tc_input, tc_max, config_reiter, excel_pfad):
    if abs(tc_max) < 2:
        return []

    if not os.path.isfile(excel_pfad):
        return []

    tc_excel = get_fnt_testcases(excel_pfad, config_reiter)
    intervalls, kats, notkats = decode_tc_selection(tc_input)
    print tc_excel
    
    tc_excel_ok = []
    tc_codierung_ok = []

    for id_, katlist, cod in tc_excel:
        for kat in katlist:
            if kat in kats:
                tc_excel_ok.append(id_)
                if cod:
                    tc_codierung_ok.append(id_)
                break
            elif notkats and kat not in notkats:
                tc_excel_ok.append(id_)
                if cod:
                    tc_codierung_ok.append(id_)
                break
        else:
            if notkats and not katlist:
                tc_excel_ok.append(id_)
                if cod:
                    tc_codierung_ok.append(id_)
            elif check_tc_spec(id_, intervalls):
                tc_excel_ok.append(id_)
                if cod:
                    tc_codierung_ok.append(id_)

    temp_out = []
    cod = None
    temp = []
    if tc_max < 0:
        tc_max = len(tc_excel_ok) / abs(tc_max) + 2  # +1 Extra-Kodiertestcase +1 Rundung
        if tc_max < 2:
            return []

    for tc in tc_excel_ok:
        if len(temp) < tc_max:
            temp.append(tc)
            if tc in tc_codierung_ok:
                cod = tc
        else:
            temp_out.append(temp)
            temp = [tc]
            if tc in tc_codierung_ok:
                cod = tc
            elif cod is not None:
                temp = [cod, tc]
    temp_out.append(temp)

    output = []
    for elem in temp_out:
        o = ""
        for e in elem:
            o += str(e) + ";"
        output.append(o[:-1])
    return output


if __name__ == "__main__":
    print testcasefilter("*", 3, "Konfiguration", r"C:\Data\DSPLIT.xlsx")

结果输出应该像是因为我们有一个pev一个phev，而第三个xml不是bev或不是phev ['1000000; 1001001'，'1003000; 1003001; 1003002'，'1004000; 1004001; 1004002']

Answer 1

这是解决方案：

test = [(1000000, ['FA'], 'bev.xml'),
 (1001001, ['TC', 'CT03', 'CT04'], False),
 (1003000, ['FA'], 'phev.xml'),
 (1003001, ['TC', 'CT01', 'CT03', 'CT04'], False),
 (1003002, ['TC', 'CT01', 'CT03', 'CT04'], False),
 (1004000, ['FA'], 'tesla.xml'),
 (1004001, ['TC', 'CT03', 'CT04'], False),
 (1004002, ['TC', 'CT03', 'CT04'], False)]

bev = [1003000]
notbev = [1000000,100400]
arrayall = [1000000,1001001,1003000,1003001,1003002,1004000,1004001,1004002]
arr_part_bev = []
arr_part_not_bev = []
isbev = False
for i in arrayall:
  if i in bev:
    arr_part_bev.append(i)
    isbev = True
  elif i in notbev:
    arr_part_not_bev.append(i)
    isbev = False  
  else:
    if isbev:
      arr_part_bev.append(i)
    else :
      arr_part_not_bev.append(i)

print (arr_part_bev)
print (arr_part_not_bev)

根据条件将大数组拆分为小部分

1 个答案: