在数据框中的每一行中获取前 n 个值和它们出现的列的名称

Question

我有一个像这样的数据框：

df = pd.DataFrame({'a':[1,2,1],'b':[4,6,0],'c':[0,4,8]})
+---+---+---+
| a | b | c |
+---+---+---+
| 1 | 4 | 0 |
+---+---+---+
| 2 | 6 | 4 |
+---+---+---+
| 1 | 0 | 8 |
+---+---+---+

对于每一行，我需要（两者）'n'（在本例中为两个）最高值和相应的降序排列的列：

row 1: 'b':4,'a':1
row 2: 'b':6,'c':4
row 3: 'c':8,'a':1

Answer 1

这里有两种方式，都是改编自

1) 使用 Python 修饰-排序-取消修饰，在每一行上加上 .apply(lambda ...) 插入列名，执行 np.argsort，保留顶部-n，重新格式化答案。（我觉得这样更干净）。

import numpy as np

# Apply Decorate-Sort row-wise to our df, and slice the top-n columns within each row...

sort_decr2_topn = lambda row, nlargest=2:
    sorted(pd.Series(zip(df.columns, row)), key=lambda cv: -cv[1]) [:nlargest]

tmp = df.apply(sort_decr2_topn, axis=1)

0    [(b, 4), (a, 1)]
1    [(b, 6), (c, 4)]
2    [(c, 8), (a, 1)]

# then your result (as a pandas DataFrame) is...
np.array(tmp)
array([[('b', 4), ('a', 1)],
       [('b', 6), ('c', 4)],
       [('c', 8), ('a', 1)]], dtype=object)
# ... or as a list of rows is
tmp.values.tolist()
#... and you can insert the row-indices 0,1,2 with 
zip(tmp.index, tmp.values.tolist())
[(0, [('b', 4), ('a', 1), ('c', 0)]), (1, [('b', 6), ('c', 4), ('a', 2)]), (2, [('c', 8), ('a', 1), ('b', 0)])]

2) 如下获取topnlocs的矩阵，然后用它重新索引到df.columns和df.values，并合并输出:

import numpy as np

nlargest = 2
topnlocs = np.argsort(-df.values, axis=1)[:, 0:nlargest]
# ... now you can use topnlocs to reindex both into df.columns, and df.values, then reformat/combine them somehow
# however it's painful trying to apply that NumPy array of indices back to df or df.values,

参见

在数据框中的每一行中获取前 n 个值和它们出现的列的名称

Get both the top-n values and the names of columns they occur in, within each row in dataframe

python

pandas

dataframe

top-n