使用字符串值在数据框中追加列

Question

我有一个案例，我试图以可理解的格式将计算出的百分比值附加到我的数据框中名为 df 的列中。当我说可理解的格式时，列中的输出应该类似于'40% Matched'，如下例所示。

df = pd.DataFrame({ 'Col1':[['Phone', 'Watch', 'Pen', 'Pencil', 'Knife'],['apple','orange','mango','cherry','banana','kiwi','tomato','avocado']], 'Col2': [['Phone', 'Watch', 'Pen', 'Pencil', 'fork'],['orange','avocado','kiwi','mango','grape','lemon','tomato']]})

df['Matched Percent'] = 'No Match'

for index,(lst1,lst2) in enumerate(zip(df['Col1'],df['Col2'])):
   if(lst1 == lst2):
      print('100% Matched')
   else:
      c1 = Counter(lst1)
      c2 = Counter(lst2)
      matching = {k: c1[k]+c2[k] for k in c1.keys() if k in c2}
      text = '% Matched'
      if len(lst1) > len(lst2):
         out = round(len(matching)/len(lst1)*100)
         #df['Matched Percent'].append(out,'% Matched')
         print(out,'% Matched')
      else:
         out = round(len(matching)/len(lst2)*100)
         #df['Matched Percent'].append(out,'% Matched')
         print(out,'% Matched')

80 % Matched
62 % Matched

TypeError: cannot concatenate object of type "<class 'int'>"; only pd.Series, pd.DataFrame, and pd.Panel (deprecated) objs are valid

我不断收到 TypeError。我尝试了几种方法，但没有运气。如上所示，我能够以我想要的方式在屏幕上打印值。但是当我将它附加到我的数据框df 时，它失败了。感谢有关如何解决此问题的建议。

Answer 1

您的逻辑似乎很冗长。您可以使用列表推导：

zipper = zip(map(set, df['Col1']), map(set, df['Col2']))
df['Matched Percent'] = [len(c1 & c2) / max(len(c1), len(c2)) for c1, c2 in zipper]

print(df)

                                                Col1  \
0                 [Phone, Watch, Pen, Pencil, Knife]   
1  [apple, orange, mango, cherry, banana, kiwi, t...   

                                                Col2  Matched Percent  
0                  [Phone, Watch, Pen, Pencil, fork]            0.800  
1  [orange, avocado, kiwi, mango, grape, lemon, t...            0.625  

请注意，使用 Pandas 优化此类计算的余地不大，Pandas 的设计目的不是用于保存串联列表。如果你需要“漂亮”的输出，你可以使用 Python 3.6+ 支持的 f-strings：

print((df['Matched Percent']*100).map(lambda x: f'{x:.0f}% Matched'))

0    80% Matched
1    62% Matched
Name: Matched Percent, dtype: object

Answer 2

如果我正确理解您的问题，

df = pd.DataFrame({ 'Col1':[['Phone', 'Watch', 'Pen', 'Pencil', 'Knife'],
                            ['apple','orange','mango','cherry','banana','kiwi','tomato','avocado']], 
                      'Col2': [['Phone', 'Watch', 'Pen', 'Pencil', 'fork'],
                               ['orange','avocado','kiwi','mango','grape','lemon','tomato']]})

df['Matched Percent'] = 'No Match'

for index,(lst1,lst2) in enumerate(zip(df['Col1'],df['Col2'])):
    if(lst1 == lst2):
        print('100% Matched')
        df['Matched Percent'][index] = '{}% Matched'.format(100)
    else:
        c1 = Counter(lst1)
        c2 = Counter(lst2)
        matching = {k: c1[k]+c2[k] for k in c1.keys() if k in c2}
        text = '% Matched'
        if len(lst1) > len(lst2):
            out = round(len(matching)/len(lst1)*100)
            #df['Matched Percent'].append(out,'% Matched')
            print(out,'% Matched')
            df['Matched Percent'][index] = '{}% Matched'.format(out)
        else:
            out = round(len(matching)/len(lst2)*100)
            #df['Matched Percent'].append(out,'% Matched')
            print(out,'% Matched')
            df['Matched Percent'][index] = '{}% Matched'.format(out)

我希望您的 df 满足以下条件：

import pandas as pd

result = 'blablabla'
df = pd.DataFrame(data=[[1,2,3],[4,5,6]],
                  columns=['a','b','Match Percent'])
s = pd.Series(index=df.columns)
df = df.append(s, ignore_index=True)
df['Match Percent'][df.shape[0]-1] = result

print(df)

     a    b Match Percent
0  1.0  2.0             3
1  4.0  5.0             6
2  NaN  NaN     blablabla

我想这就是你想要的，对吧？

更正：

df = pd.DataFrame(data=[[1,2],[4,5]],
              columns=['a','b'])

df['Match Percent'] = (df['a']/df['b']*100).apply(lambda x: '{} Match'.format(x))


   a  b Match Percent
0  1  2    50.0 Match
1  4  5    80.0 Match