Question

我有一个这样的数据框：

         desc     id     info  
       [a,b,c]     2     type
       [u,v,w]     18    tail

三列：desc，id，info和desc是一个列表。我想要这个：

        des    id    info 
         a      2     type
         b      2     type
         c      2     type 
         u      18    tail
         v      18    tail
         w      18    tail

这意味着将列表列分解为多行，其他列没有更改。我真的不知道该怎么做......

Answer 1

这是一种方式

df.set_index(['id', 'info']).desc.apply(pd.Series).stack()\
.reset_index(name = 'desc').drop('level_2', axis = 1)


    id  info    desc
0   2   type    a
1   2   type    b
2   2   type    c
3   18  tail    u
4   18  tail    v
5   18  tail    w

Answer 2

我记得这应该来自piRSquared或cᴏʟᴅsᴘᴇᴇᴅ，但找不到链接...

function index($type="deposit", $limit_from=0)
{
    $limit_from = ($this->uri->segment(4)) ? $this->uri->segment(4) : 0;
    $total_records = $this->Expense->get_total_rows($type);
    $data['controller_name'] = $this->get_controller_name();

    $lines_per_page = 3; 
    $data[$type.'s'] = $this->Expense->get_records($lines_per_page, $limit_from, $type);

    $config['base_url'] = base_url('index.php/expenses/index/'.$type);
    $config['total_rows'] = $total_records;
    $config['per_page'] = 3;
    $config["uri_segment"] = 4;
    // $choice = $total_records / $config['per_page'];
    // $config['num_links'] = round($choice);
    $config['use_page_numbers'] = TRUE;
    $config['first_url'] = base_url('index.php/expenses/index/'.$type.'/1');

    $this->pagination->initialize($config);
    $data["links"] = $this->pagination->create_links();

    $this->load->view('expenses/'.$type, $data);
    $this->_remove_duplicate_cookies();
}

Answer 3

您可以展平desc列，repeat其他两列，然后将它们连接起来：

pd.concat([
    pd.Series([e for s in df.desc for e in s], name='desc'),
    df.drop('desc', 1).apply(lambda col: col.repeat(df.desc.str.len())).reset_index(drop=True)
], axis=1)

#desc   id  info
#0  a    2  type
#1  b    2  type
#2  c    2  type
#3  u   18  tail
#4  v   18  tail
#5  w   18  tail

Answer 4

你可以

In [1631]: (df.loc[df.index.repeat(df.desc.str.len())]
              .assign(desc=[v for x in df.desc.values for v in x]))
Out[1631]:
  desc  id  info
0    a   2  type
0    b   2  type
0    c   2  type
1    u  18  tail
1    v  18  tail
1    w  18  tail

Python数据帧将列表列分解为多行

4 个答案: