在每一行中查找最接近的列值 - 熊猫

Question

以下是更大数据集的示例：

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]})

我预测了需要与观测值进行比较的风速......最终我需要找到最接近观测风速值的预测速度（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

制作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

这是实现您正在寻找的另一种方法。 它允许比较多于两列。

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

修改杰夫的解决方案我设法想出了这个：

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某些列/行中的字符串不是“NaN”，因此我使用了 Pandas 并强制执行了错误：

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')

使用杰夫的建议应用功能：

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

可能不是最有效的，但我完成了工作......感谢大家的帮助。