如何从 df 中删除 nan 而不丢失整行？

Question

G'day，如何在不丢失整行的情况下删除 nan 值？这就是我的样子。df

我已经尝试了但这删除了行中的所有内容。pandasdf = schule.dropna()

但我必须保持这些价值观，因为在最后，我希望它们向上发展。

Answer 1

它可能会有所帮助。你可以尝试一下!!!!

Answer 2

根据问题陈述，您希望降低 nan 值的优先级，并将非 nan 值置于顶部。import numpy as npimport pandas as pdimport functoolsdef drop_and_roll(col, na_position='last', fillvalue=np.nan):    result = np.full(len(col), fillvalue, dtype=col.dtype)    mask = col.notnull()    N = mask.sum()    if na_position == 'last':        result[:N] = col.loc[mask]    elif na_position == 'first':        result[-N:] = col.loc[mask]    else:        raise ValueError('na_position {!r} unrecognized'.format(na_position))    return resultdf = pd.read_table('data', sep='\s{2,}')print(df.apply(functools.partial(drop_and_roll, fillvalue='')))

Answer 3

假设您要回填值，然后删除任何列中显示的任何重复项，此示例有效：import pandas as pdimport numpy as npdata = [&nbsp; &nbsp; ['POINT_1.1', 'POINT_1.2', pd.NA],&nbsp; &nbsp; [pd.NA, pd.NA, 'POINT_1.3'],&nbsp; &nbsp; ['POINT_2.1', 'POINT_2.2', pd.NA],&nbsp; &nbsp; [pd.NA, pd.NA, 'POINT_2.3']]df = pd.DataFrame(data)df#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2# 0&nbsp; POINT_1.1&nbsp; POINT_1.2&nbsp; &nbsp; &nbsp; &nbsp;<NA># 1&nbsp; &nbsp; &nbsp; &nbsp;<NA>&nbsp; &nbsp; &nbsp; &nbsp;<NA>&nbsp; POINT_1.3# 2&nbsp; POINT_2.1&nbsp; POINT_2.2&nbsp; &nbsp; &nbsp; &nbsp;<NA># 3&nbsp; &nbsp; &nbsp; &nbsp;<NA>&nbsp; &nbsp; &nbsp; &nbsp;<NA>&nbsp; POINT_2.3t = df.T.bfill().T.bfill()t#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2# 0&nbsp; POINT_1.1&nbsp; POINT_1.2&nbsp; POINT_1.3# 2&nbsp; POINT_2.1&nbsp; POINT_2.2&nbsp; POINT_2.3for column in t.columns:&nbsp; &nbsp; t = t.drop_duplicates(column)t#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2# 0&nbsp; POINT_1.1&nbsp; POINT_1.2&nbsp; POINT_1.3# 2&nbsp; POINT_2.1&nbsp; POINT_2.2&nbsp; POINT_2.3