Question

我已经用谷歌搜索了它，通常的解决方案似乎是，迭代器必须是常量。但我仍然得到这个错误。这是我的代码：

std::vector<std::vector<std::vector<std::string> > >::const_iterator itc;
std::vector<std::vector<std::string> >::const_iterator itb;
std::vector<std::string>::const_iterator ita;
for (ita = metadata.getTestVektor().begin(); ita != metadata.getTestVektor().end(); ++ita)
{
    for (itb = ita->begin(); itb != ita->end(); ++itb)
    {
        for (itc = itb->begin(); itc != itb->end(); ++itc)
        {
        }
    }
}

metadata.getTestVektor（）返回3d vektor：

std::vector<std::vector<std::vector<std::string>>> testvektor;

日志：

/home/MetaDataCreator.cpp:20: error: no match for 'operator=' (operand types are 'std::vector<std::basic_string<char> >::const_iterator {aka __gnu_cxx::__normal_iterator<const std::basic_string<char>*, std::vector<std::basic_string<char> > >}' and 'std::vector<std::vector<std::vector<std::basic_string<char> > > >::iterator {aka __gnu_cxx::__normal_iterator<std::vector<std::vector<std::basic_string<char> > >*, std::vector<std::vector<std::vector<std::basic_string<char> > > > >}') for (ita = metadata.getTestVektor().begin(); ita != metadata.getTestVektor().end(); ++ita) ^

^点在“=”符号

Answer 1

好像你已经混淆了In [91]: merged = pd.merge(df1,df2,indicator=True, how='left' ) merged Out[91]: RecorderID GroupID Location SomeColumn _merge 0 CT-1000001 BV- Cape Town SomeValue left_only 1 CT-1000002 MP- Johannesburg SomeValue left_only 2 CT-1000003 BV- Durban SomeValue both In [92]: merged[merged['_merge'] == 'left_only'] Out[92]: RecorderID GroupID Location SomeColumn _merge 0 CT-1000001 BV- Cape Town SomeValue left_only 1 CT-1000002 MP- Johannesburg SomeValue left_only和ita，其中itc具有最里面的迭代器类型。

但也要考虑Joachim所说的内容 - 如果ita按值返回，则在两个不同的向量上调用getTestVektor()和begin()。不好！</ p>

C ++迭代3d数组'不匹配operator ='

1 个答案: