在每行中查找最接近的列值- pandas

Question

以下是一个更大的数据集的示例：

df_old = pd.DataFrame({'code': ['fea-1','fea-132','fea-223','fea-394','fea-595','fea-130','fea-495'],
                   'forecastWind_low':[20,15,0,45,45,25,45],
                   'forecastWind_high':['NaN' ,30,'NaN',55,65,35,'NaN'],
                   'obs_windSpeed':[20,11,3,65,55,'NaN',55]})

我预测了风速，我需要与观测值进行比较。最终，我需要找到与观测风速值最接近的预测风速(低或高)，以获得如下输出：

df_new = pd.DataFrame({'code': ['fea-1','fea-132','fea-223','fea-394','fea-595','fea-130','fea-495'],
                   'forecastWind_low':[20,15,0,45,45,25,45],
                   'forecastWind_high':['NaN' ,30,'NaN',55,65,35,'NaN'],
                   'obs_windSpeed':[20,11,3,65,55,'NaN',55],
                   'nearest_forecast_windSpeed':[20,15,0,55,45,'NaN',45]})

Answer 1

创建自定义比较函数并将其应用于各行

def check_speed_diff(high,low,obs):
    if np.isnan(obs):
        return np.nan
    elif np.isnan(high):
        return low
    elif np.isnan(low):
        return high
    
    if abs(high-obs)<abs(low-obs):
        return high
    else:
        return low

df_old.apply(lambda x: 
    check_speed_diff(
        x.forecastWind_high,
        x.forecastWind_low,
        x.obs_windSpeed
    ),
    axis=1
)

Answer 2

这是另一种实现你想要的东西的方法。它允许不止两列进行比较。

col = ['forecastWind_low','forecastWind_high']
comparecol = ['obs_windSpeed']
df[col + comparecol] = df[col + comparecol].astype(float)
dfmerge =pd.merge(df[col].stack().reset_index(-1),df[comparecol],left_index=True,right_index=True,how='left')
dfmerge = dfmerge.rename(columns = {'level_1':'windforecast',0:'Amount'})
dfmerge['difference'] = abs(dfmerge['obs_windSpeed'] - dfmerge['Amount'])
dfmerge = dfmerge.sort_values(by='difference',ascending=True)
dfmerge = dfmerge.groupby(level=0).head(1)
df = pd.merge(df,dfmerge['Amount'],left_index=True,right_index=True,how='left')
df.loc[df['obs_windSpeed'].isna(),'Amount'] = np.nan

Answer 3

在修改Jeff的解决方案时，我设法想出了这个：

def check_speed_diff(high,low,obs):
    if obs == 'NaN':
        return np.nan
    if low != 'NaN' and high == 'NaN':
        return low
    if low == 'NaN' and high != 'NaN':
        return high
    if low != 'NaN' and high != 'NaN':
        if abs(high-obs)<abs(low-obs):
            return high
        else:
            return low

我遇到的另一个问题是一些列/行中的字符串不是'NaN'，所以我使用pandas并强制执行错误：

df.forecast_WindSpeed_high = pd.to_numeric(df.forecast_WindSpeed_high,errors='coerce')
df.forecast_WindSpeed_low = pd.to_numeric(df.forecast_WindSpeed_low ,errors='coerce')

使用Jeff建议的应用函数：

df['nearest_forecastWindSpeed'] = df.apply(lambda x: check_speed_diff(
        x.forecast_WindSpeed_high, 
        x.forecast_WindSpeed_low,
        x.windSpeed),axis=1)

可能不是最有效率的，但我完成了工作...感谢大家的帮助。