Question

我有一个这样的数据框，

df
col1    col2    col3
1907    CD       49
1907    FR       33
1907    SA       34
1908    PR        1
1908    SA       37
1909    PR       16
1909    SA       38

现在CD不具有col1 1908和1909值，FR不具有1908和1909值，PR不存在1907。

现在，我想创建具有col2值的行，而这些行的col3值不等于0的所有col1值。

所以最终的数据帧看起来像

df
col1    col2    col3
1907    CD       49
1907    FR       33
1907    SA       34
1907    PR        0
1908    CD        0
1908    FR        0
1908    PR        1
1908    SA       37
1908    CD        0
1908    FR        0
1909    PR       16
1909    SA       38

我可以使用带有每个可能col2值的for循环并与每个col1组进行比较来做到这一点。但是我正在寻找最有效的捷径。

Answer 1

将DataFrame.unstack与DataFrame.stack一起用于由var isFetching: Bool = false var offset = 0 var totalListOnServerCount = 20 // it must be returned from server var pageSize = 10 // get 10 objects for instance // MARK: - API Handler private func fetchNotifications(){ // return from function if already fetching list guard !isFetching else {return} if offset == 0{ // empty list for first call i.e offset = 0 self.anyList.removeAll() self.collectionView.reloadData() } isFetching = true // API call to fetch notifications with given offset or page number depends on server logic just simple POST Call APIHandler.shared.getNotifications(offset: offset) {[weak self] (response, error) in if let response = response { self?.isFetching = false if self?.offset == 0{ // fetch response from server for first fetch self?.notificationsResponse = response if self?.refreshControl.isRefreshing ?? false { self?.refreshControl.endRefreshing() } }else{ // append if already exist ( pagination ) self?.notificationsResponse?.notifications.append(contentsOf: response.notifications) } self?.collectionView.reloadData() } } } // MARK: - Collection View Delegate func collectionView(_ collectionView: UICollectionView, willDisplay cell: UICollectionViewCell, forItemAt indexPath: IndexPath) { guard let anyList = responseFromServer else { return } // check if scroll reach last index available and keep fetching till our model list has all entries from server if indexPath.item == anyList.count - 1 && anyList.count < totalListOnServerCount{ offset += pageSize fetchNotifications() } }填充的所有组合：

另一个想法是将DataFrame.reindex与MultiIndex.from_product结合使用：

df = df.set_index(['col1','col2']).unstack(fill_value=0).stack().reset_index()
print (df)
    col1 col2  col3
0   1907   CD    49
1   1907   FR    33
2   1907   PR     0
3   1907   SA    34
4   1908   CD     0
5   1908   FR     0
6   1908   PR     1
7   1908   SA    37
8   1909   CD     0
9   1909   FR     0
10  1909   PR    16
11  1909   SA    38

Answer 2

我们还可以对DataFrame.stack做DataFrame.pivot_table：

df.pivot(*df).stack(dropna = False).fillna(0).rename('col3').reset_index()

或{{3}}

df.pivot_table(*df.iloc[:,::-1],fill_value = 0).unstack().rename('col3').reset_index()

输出

    col1 col2  col3
0   1907   CD  49.0
1   1907   FR  33.0
2   1907   PR   0.0
3   1907   SA  34.0
4   1908   CD   0.0
5   1908   FR   0.0
6   1908   PR   1.0
7   1908   SA  37.0
8   1909   CD   0.0
9   1909   FR   0.0
10  1909   PR  16.0
11  1909   SA  38.0

用熊猫数据框中的一些默认值填充每个列组合的值

2 个答案: