我正在關注這個問題的答案
我有一個df這樣的：
score_1 score_2
1.11 NaN
2.22 3.33
NaN 3.33
NaN NaN
........
計算規則final_score是我們要求至少有一個分數為non-null，如果其中一個分數為 NULL，則 final_score 將等于另一個分數（它具有所有權重） 這是要復制的代碼：
import numpy as np
import pandas as pd
df = pd.DataFrame({
'score_1': [1.11, 2.22, np.nan],
'score_2': [np.nan, 3.33, 3.33]
})
def final_score(df):
if (df['score_1'] != np.nan) and (df['score_2'] != np.nan):
print('I am condition one')
return df['score_1'] * 0.2 df['score_2'] * 0.8
elif (df['score_1'] == np.nan) and (df['score_2'] != np.nan):
print('I am the condition two')
return df['score_2']
elif (df['score_1'] != np.nan) and (df['score_2'] == np.nan):
print('I am the condition three')
return df['score_1']
elif (df['score_1'] == np.nan) and (df['score_2'] == np.nan):
print('I am the condition four')
return np.nan
df['final_score'] = df.apply(final_score, axis=1)
print(df)
這給了我輸出：
score_1 score_2 final_score
1.11 NaN NaN
2.22 3.33 3.108
NaN 3.33 NaN
NaN NaN NaN
........
但我的預期輸出如下：
score_1 score_2 final_score
1.11 NaN 1.11
2.22 3.33 3.108
NaN 3.33 3.33
NaN NaN NaN
........
第一行和第三行不是我期望的結果，有人可以幫助我，我的代碼有什么問題嗎？非常感謝。
uj5u.com熱心網友回復：
讓我們使用 np.where 應用你的條件
df['final_score'] =np.where(df.notna().all(1),df['score_1'] * 0.2 df['score_2'] * 0.8,df.mean(1))
score_1 score_2 final_score
0 1.11 NaN 1.110
1 2.22 3.33 3.108
2 NaN 3.33 3.330
3 NaN NaN NaN
uj5u.com熱心網友回復：
使用 np.isnan() 進行比較應該可以解決問題
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