根據在另一列中設置的金額向零和列添加零點

Question

我有一列包含電話號碼（字符串）。然後有另一列給了我零的數量（如int），我想追加到現有的電話號碼。我想要做的是像（5 *「0」=>「00000」）。還有一個先決條件。如果電話號碼結束於「1」，則只能添加零。

Example: 
>>> df = pd.DataFrame([["11", 2], ["21", 3], ["10", 6]], columns=['phone', 'ext_length']) 

What I tried: 
>>> df.loc[(df.phone.str.endswith("1")), "complete_phone"] = df.phone + (df.ext_length * "0") 

同時過濾正確的行，其中的手機上「1」結尾，創造列「complete_phone」的作品，我不能得到的「數學」的工作。我越來越

TypeError: ufunc 'multiply' did not contain a loop with signature matching types dtype('<U21') dtype('<U21') dtype('<U21') 

我也不理解錯誤消息，況且我有一個想法，如何解決這個問題。我還在尋找一個鏈接，它顯示瞭如何正確包含Python示例，正如我在[in：]和[out：]加結果中看到的其他問題。任何提示？

Answer 1

我認爲你需要mask的更換，如果條件True和 str.repeat：與DataFrame.apply

s = pd.Series(['0'], index=df.index) 
mask = df.phone.str.endswith("1") 
df["complete_phone"] = df.phone.mask(mask, df.phone + s.str.repeat(df.ext_length)) 
print (df) 
    phone ext_length complete_phone 
0 11   2   1100 
1 21   3   21000 
2 10   6    10 

另一種解決方案：

mask = df.phone.str.endswith("1") 
df["complete_phone"] = df['phone'].mask(mask, df.apply(lambda x: x['phone'] + 
                   '0' * x.ext_length, axis=1)) 
print (df) 
    phone ext_length complete_phone 
0 11   2   1100 
1 21   3   21000 
2 10   6    10 

mask = df.phone.str.endswith("1") 
df["complete_phone"] = df.phone.mask(mask, df['phone'] + 
              df['ext_length'].apply(lambda x:'0'*x)) 
print (df) 
    phone ext_length complete_phone 
0 11   2   1100 
1 21   3   21000 
2 10   6    10 

你的解決方案是類似的只得到NaN如果馬SK是False：

mask = df.phone.str.endswith("1") 
df.loc[mask, "complete_phone"] = df['phone'] + df.apply(lambda x: '0' * x.ext_length, axis=1) 
    phone ext_length complete_phone 
0 11   2   1100 
1 21   3   21000 
2 10   6   NaN 

Answer 2

In [1]: import pandas as pd 

In [2]: df = pd.DataFrame([["11", 2], ["21", 3], ["10", 6]], columns=['phone', 'ext_length']) 

In [3]: df 
Out[3]: 
    phone ext_length 
0 11   2 
1 21   3 
2 10   6 

In [4]: df['complete_phone'] = [number+'0'*length if number.endswith('1') else number for number, length in zip(df.phone, df.ext_length)] 

In [5]: df 
Out[5]: 
    phone ext_length complete_phone 
0 11   2   1100 
1 21   3   21000 
2 10   6    10 

Answer 3

我覺得apply功能就派上用場了：

df = pd.DataFrame([["11", 2], ["21", 3], ["10", 6]], columns=['phone', 'ext_length']) 
df['complete_phone'] = df.apply(lambda x: x['phone'] + "0" * x['ext_length'], axis=1)