与您链接的答案一样，您希望创建一个标识组编号的列。然后，您可以使用相同的解决方案。
为此，您必须测试一行的所有值是否为NaN。我不知道pandas中是否有这样一个内置的测试，但是pandas有一个测试来检查一个Series是否充满了NaN。所以你要做的是对 Dataframe 的转置执行，这样你的“Series”实际上就是你的行：

df["group_no"] = df.isnull().all(axis=1).cumsum()

在这一点上，你可以使用相同的技术，从该答案分裂的边框。
您可能希望在最后执行.dropna()，因为结果中仍然有NaN行。

在2022年遇到了同样的问题。下面是我用NaN在行上分割行的方法，需要注意的是，这依赖于pip install python-rle的游程编码：

import rle
def nanchucks(df):
    # It chucks NaNs outta dataframes
    
    # True if whole row is NaN
    df_nans = pd.isnull(df).sum(axis="columns").astype(bool)
    values, counts = rle.encode(df_nans)
    
    df_nans = pd.DataFrame({"values": values, "counts": counts})
    df_nans["cum_counts"] = df_nans["counts"].cumsum()
    df_nans["start_idx"] = df_nans["cum_counts"].shift(1)
    df_nans.loc[0, "start_idx"] = 0
    df_nans["start_idx"] = df_nans["start_idx"].astype(int) # np.nan makes it a float column
    df_nans["end_idx"] = df_nans["cum_counts"] - 1
    
    # Only keep the chunks of data w/o NaNs
    df_nans = df_nans[df_nans["values"] == False]
    
    indices = []
    for idx, row in df_nans.iterrows():
        indices.append((row["start_idx"], row["end_idx"]))
        
    return [df.loc[df.index[i[0]]: df.index[i[1]]] for i in indices]

示例如下：

sample_df1 = pd.DataFrame({
    "a": [1, 2, np.nan, 3, 4],
    "b": [1, 2, np.nan, 3, 4],
    "c": [1, 2, np.nan, 3, 4],
})
sample_df2 = pd.DataFrame({
    "a": [1, 2, np.nan, 3, 4],
    "b": [1, 2, 3, np.nan, 4],
    "c": [1, 2, np.nan, 3, 4],
})
print(nanchucks(sample_df1))
# [     a    b    c
#  0  1.0  1.0  1.0
#  1  2.0  2.0  2.0,
#       a    b    c
#  3  3.0  3.0  3.0
#  4  4.0  4.0  4.0]
print(nanchucks(sample_df2))
# [     a    b    c
# 0  1.0  1.0  1.0
# 1  2.0  2.0  2.0,
#       a    b    c
# 4  4.0  4.0  4.0]

使用NaNs改进其他支持多行的答案：

from IPython.display import display
import pandas as pd
def split_df_if_row_full_nans(df, reset_header=False):
    # grouping
    df = (df
          .assign(_nan_all_cols=df.isnull().all(axis=1))
          .assign(_group_no=lambda df_: df_._nan_all_cols.cumsum())
          .query('_nan_all_cols == False')  # Drop rows where _nan_all_cols is True
          .drop(columns=['_nan_all_cols'])  # Drop the _nan_all_cols column
          .reset_index(drop=True)
         )
    # splitting
    dfs = {df.iloc[rows[0],0]: (df
                                .iloc[rows]
                                .drop(columns=['_group_no'])
                                )
           for _, rows in df.groupby('_group_no').groups.items()}
    if reset_header:
        # rename column and set index
        for k, v in dfs.items():
            dfs[k] = (v
                    .rename(columns=v.iloc[0])
                    .drop(index=v.index[0])
                    )
            # TODO: this part seems to only works if length of the df is > 1
            # dfs[k].set_index(dfs[k].columns[0], drop=True, inplace=True)
    # # display
    # for df in dfs.values():
    # display(df)
    return dfs
sample_df1 = pd.DataFrame({
    "a": [1, 2, np.nan, 3, 4],
    "b": [1, 2, np.nan, 3, 4],
    "c": [1, 2, np.nan, 3, 4],
})
sample_df2 = pd.DataFrame({
    "a": [1, 2, np.nan, 3, 4],
    "b": [1, 2, 3, np.nan, 4],
    "c": [1, 2, np.nan, 3, 4],
})
for df in split_df_if_row_full_nans(sample_df1).values():
    display(df)
#   1.0 1.0 1.0
# 1 2   2   2
#     3.0   3.0 3.0
# 3 4   4   4
for df in split_df_if_row_full_nans(sample_df2).values():
    display(df)
#     1.0   1.0 1.0
# 1 2   2   2
# 2 NaN 3   NaN
# 3 3   NaN 3
# 4 4   4   4

注意：此方法使用.isnull().all(axis=1)，即仅当所有值都是NaN时才进行拆分。