Question

我有一个我从数据框创建的矩阵，我想删除其中每个值都为0的所有列。

我见过使用dropna df2.loc[:, (df2 != 0).any(axis=0)]的例子，但它对我的数据帧没有任何作用。

这就是我创建矩阵的方式：

a = ['Psychology','Education','Social policy','Sociology','Pol. sci. & internat. studies','Development studies','Social anthropology','Area Studies','Science and Technology Studies','Law & legal studies','Economics','Management & business studies','Human Geography','Environmental planning','Demography','Social work','Tools, technologies & methods','Linguistics','History']
final_df = new_df[new_df['Subject'].isin(a)]

ctrs = {location: Counter(gp.GrantRefNumber) for location, gp in final_df.groupby('Subject')}

ctrs = list(ctrs.items())
overlaps = [(loc1, loc2, sum(min(ctr1[k], ctr2[k]) for k in ctr1))
    for i, (loc1, ctr1) in enumerate(ctrs, start=1)
    for (loc2, ctr2) in ctrs[i:] if loc1 != loc2]
overlaps += [(l2, l1, c) for l1, l2, c in overlaps]

df22 = pd.DataFrame(overlaps, columns=['Loc1', 'Loc2', 'Count'])
df22 = df22.set_index(['Loc1', 'Loc2'])
df22 = df22.unstack().fillna(0).astype(int)

#the end part of the next line filters the top 'x' amount.
b = np.sort(np.unique(df22.values.ravel()))[-20:]
df2 = df22.where(df22.isin(b),0.0)

有趣的是（或不是），当我输入df2.columns时，我得到：

MultiIndex(levels=[[u'Count'], [u'Area Studies', u'Demography', u'Development studies', u'Economics', u'Education', u'Environmental planning', u'History', u'Human Geography', u'Law & legal studies', u'Linguistics', u'Management & business studies', u'Pol. sci. & internat. studies', u'Psychology', u'Science and Technology Studies', u'Social anthropology', u'Social policy', u'Social work', u'Sociology', u'Tools, technologies & methods']],
           labels=[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]],
           names=[None, u'Loc2'])

这可能就是我挣扎的原因。

Answer 1

True列的0需要all，其中包含~和df = pd.DataFrame({'B':[0,0,0,0,0,0], 'C':[0,8,9,4,2,3], 'D':[0,3,5,7,1,0], 'E':[0,3,6,9,2,4]}) print (df) B C D E 0 0 0 0 0 1 0 8 3 3 2 0 9 5 6 3 0 4 7 9 4 0 2 1 2 5 0 3 0 4 df = df.loc[~df.eq(0).all(axis=1), ~df.eq(0).all()] print (df) C D E 1 8 3 3 2 9 5 6 3 4 7 9 4 2 1 2 5 3 0 4的反转条件：

{{1}}

Python - 删除Matrix中的行和列，其中所有值均为0

1 个答案: