python - Pandas 数据框条件替换和列修剪

Question

Current Pandas DataFrame

fn1 = pd.DataFrame([['A', 'NaN', 'NaN', 9, 6], ['B', 'NaN', 2, 'NaN', 7], ['C', 3, 2, 'NaN', 10], ['D', 'NaN', 7, 'NaN', 'NaN'], ['E', 'NaN', 'NaN', 3, 3], ['F', 'NaN', 'NaN', 7,'NaN']], columns = ['Symbol', 'Condition1','Condition2', 'Condition3', 'Condition4'])

fn1.set_index('Symbol', inplace=True)



         Condition1 Condition2 Condition3 Condition4
Symbol                                            
A             NaN        NaN          9          6
B             NaN          2        NaN          7
C               3          2        NaN         10
D             NaN          7        NaN        NaN
E             NaN        NaN          3          3
F             NaN        NaN          7        NaN

我目前正在使用一个 Pandas DataFrame，它看起来像上面的链接。我试图逐列将非“NaN”的值替换为与该行关联的“符号”，然后折叠每列(或写入新的数据帧)，以便每列都是“符号”的列表每个“条件”都存在，如所需输出所示:

Desired Output

我已经能够将每个条件存在的“符号”放入列表列表中(见下文)，但希望保持相同的列名称，并且无法将它们添加到不断增长的新 DataFrame 中，因为长度是可变的，我循环遍历列。

ls2 = []
for col in fn1.columns:
    fn2 = fn1[fn1[col] > 0]
    ls2.append(list(fn2.index))

其中 fn1 是看起来像第一个图像的 DataFrame，我已将“Symbol”列设为索引。

预先感谢您的帮助。

Answer 1

另一个答案是切片，如下所示(评论中的解释):

import numpy as np
import pandas as pd

df = pd.DataFrame.from_dict({
    "Symbol": ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k"],
    "Condition1": [1, np.nan, 3, np.nan, np.nan, np.nan, 7, np.nan, np.nan, 8, 12],
    "Condition2": [np.nan, 2, 2, 7, np.nan, np.nan, 5, 11, 14, np.nan, np.nan],
    }
)


new_df = pd.concat(
    [
        df["Symbol"][df[column].notnull()].reset_index(drop=True) # get columns without null and ignore the index (as your output suggests)
        for column in list(df)[1:] # Iterate over all columns except "Symbols"
    ],
    axis=1, # Column-wise concatenation
)
# Rename columns
new_df.columns = list(df)[1:]
# You can leave NaNs or replace them with empty string, your choice
new_df.fillna("", inplace=True)

此操作的输出将是:

  Condition1 Condition2
0          a          b
1          c          c
2          g          d
3          j          g
4          k          h
5                     i

如果您需要任何进一步的说明，请在下面发表评论。