我有一个数据框df
列中包含浮点值A
。我想添加另一列B
这样:
-
B[0] = A[0]
for 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]
示例数据框df
可以如下生成
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
这可以相当有效地完成Numba http://numba.pydata.org/。如果您无法使用 Numba,则省略@njit
并且您的逻辑将作为 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
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