Question

我有一个数据框df，列A中有浮点值。我想添加另一列B，例如：

B[0] = A[0]

i > 0 ...
B[i] = if(np.isnan(A[i])) then A[i] else Step3
B[i] = if(abs((B[i-1] - A[i]) / B[i-1]) < 0.3) then B[i-1] else A[i]

可以如下所示生成示例数据帧df

import numpy as np
import pandas as pd
df = pd.DataFrame(1000*(2+np.random.randn(500, 1)), columns=list('A'))
df.loc[1, 'A'] = np.nan
df.loc[15, 'A'] = np.nan
df.loc[240, 'A'] = np.nan
df.loc[241, 'A'] = np.nan

Answer 1

使用Numba可以相当有效地完成此操作。如果您无法使用Numba，只需省略@njit，您的逻辑将作为Python级循环运行。

import numpy as np
import pandas as pd
from numba import njit

np.random.seed(0)
df = pd.DataFrame(1000*(2+np.random.randn(500, 1)), columns=['A'])
df.loc[1, 'A'] = np.nan
df.loc[15, 'A'] = np.nan
df.loc[240, 'A'] = np.nan

@njit
def recurse_nb(x):
    out = x.copy()
    for i in range(1, x.shape[0]):
        if not np.isnan(x[i]) and (abs(1 - x[i] / out[i-1]) < 0.3):
            out[i] = out[i-1]
    return out

df['B'] = recurse_nb(df['A'].values)

print(df.head(10))

             A            B
0  3764.052346  3764.052346
1          NaN          NaN
2  2978.737984  2978.737984
3  4240.893199  4240.893199
4  3867.557990  4240.893199
5  1022.722120  1022.722120
6  2950.088418  2950.088418
7  1848.642792  1848.642792
8  1896.781148  1848.642792
9  2410.598502  2410.598502

Answer 2

不确定第一个B-1和除以NaN的情况要做什么：

df = pd.DataFrame([1,2,3,4,5,None,6,7,8,9,10], columns=['A'])
b1 = df.A.shift(1)
b1[0] = 1
b = list(map(lambda a,b1: a if np.isnan(a) else (b1 if abs(b1-a)/b1 < 0.3 else a), df.A, b1 ))
df['B'] = b

df
       A    B
0    1.0  1.0
1    2.0  2.0
2    3.0  3.0
3    4.0  4.0
4    5.0  4.0
5    NaN  NaN
6    6.0  6.0
7    7.0  6.0
8    8.0  7.0
9    9.0  8.0
10  10.0  9.0

按照@jpp，您还可以为列表b做一个列表理解版本：

b = [a if np.isnan(a) or abs(b-a)/b >= 0.3 else b for a,b in zip(df.A,b1)]

Answer 3

下面是一个我可以想到的简单解决方案。我想知道是否还有更多的Python方式：

 a = df['A'].values
 b = []
 b.append(t[0])
 for i in range(1, len(a)):
     if np.isnan(a[i]):
         b.append(a[i])
     else:
         b.append(b[i-1] if abs(1 - a[i]/b[i-1]) < 0.3 else a[i])
 df['B'] = b

Answer 4

因此，对于现实世界的数据而言，这可能更快，但也存在真正的最坏情况（如果行0 >>其余数据，则while循环将迭代N次）。

df['B'] = df['A']
to_be_fixed = pd.Series(True, index=df.index)
while to_be_fixed.any():
    # Shift column B and the rows that need to be logically tested
    diff = df['B'].shift(1)
    to_be_fixed = to_be_fixed.shift(1)

    # Test the rows to see which need to be replaced
    to_be_fixed = to_be_fixed & (np.abs(1 - df['A'] / diff) < 0.3)

    # Replace data
    df.loc[to_be_fixed, 'B'] = diff.loc[to_be_fixed]

    # Fix np.nan that has been introduced into column B
    b_na = pd.isnull(df['B'])
    df.loc[b_na, 'B'] = df.loc[b_na, 'A']

有条件地填写熊猫数据框

4 个答案: