在由現有列中的值組成的數據框中創建新列

Question

我有一個看起來像這樣的數據框：

      X       Y   Corr_Value
  0 51182   51389   1.00
  1 51182   50014   NaN
  2 51182   50001   0.85
  3 51182   50014   NaN

我想創建一個由X和Y列的值組成的新列。 想法是遍歷行，如果Corr_Value不為null，則新列應顯示：

Solving (X column value) will solve (Y column value) at (Corr_value column)% probability.

例如，對於第一行，結果應為：

Solving 51182 will solve 51389 with 100% probability.

這是我寫的代碼：

dfs = []
for i in df1.iterrows():
    if ([df1['Corr_Value']] != np.nan):

        a = df1['X']
        b = df1['Y']
        c = df1['Corr_Value']*100
        df1['Remarks'] = (f'Solving {a} will solve {b} at {c}% probability')
        dfs.append(df1)

df1是存儲X ， Y和Corr_Value數據的數據幀。

但是似乎有一個問題，因為我得到的結果看起來像這樣：

但是結果應如下所示：

如果您可以幫助我獲得理想的結果，那就太好了。

Answer 1

使用DataFrame.dropna用於刪除丟失的行和應用f-string S表示自定義輸出字符串DataFrame.apply ：

f = lambda x: f'Solving {int(x["X"])} will solve {int(x["Y"])} at {int(x["Corr_Value"] * 100)}% probability.'
df['Remarks'] = df.dropna(subset=['Corr_Value']).apply(f,axis=1)
print (df)
       X      Y  Corr_Value                                            Remarks
0  51182  51389        1.00  Solving 51182 will solve 51389 at 100% probabi...
1  51182  50014         NaN                                                NaN
2  51182  50001        0.85  Solving 51182 will solve 50001 at 85% probabil...
3  51182  50014         NaN                                                NaN

Answer 2

您還可以在以下位置使用numpy：

import numpy as np

df['Remarks'] = np.where(df.Corr_Value.notnull(), 'Solving ' + df['X'].astype(str) + ' will solve ' + df['Y'].astype(str) + ' with ' + (df['Corr_Value'] * 100).astype(str) + '% probability', df['Corr_Value'])

輸出：

       X      Y  Corr_Value                                            Remarks
0  51182  51389        1.00  Solving 51182 will solve 51389 with 100.0% pro...
1  51182  50014         NaN                                                NaN
2  51182  50001        0.85  Solving 51182 will solve 50001 with 85.0% prob...
3  51182  50014         NaN                                                NaN

Answer 3

你試一試：

dfs = []
for i, r in df1.iterrows():
    if (r['Corr_Value'] != np.nan):
        a = r['X']
        b = r['Y']
        c = r['Corr_Value']*100
        df1.at[i, 'Remarks'] = "Solving "+  str(a) + " will solve " + str(b) + " at " + str(c) + " % probability"

我認為問題與使用df1而不是當前行有關。