Question

我正在根据其他符合条件的列创建一个新的字符串列。

我的目标是向外扩展以读取12个字段/ 30,000行错误分类的数据。

示例数据：

df = pd.DataFrame({'clothes': ['furry boots', 'weird boots', 'furry gloves', 'weird gloves', 'furry coat', 'weird coat'],
      'barn': ['furry animal', 'big animal', 'furry fence', 'old fence', 'furry door', 'old door'],
      'crazy': ['heckin food', 'furry food', 'furry toes', 'old toes', 'furry hat', 'crazy cat']})
df

+---+--------------+--------------+-------------+
|   |   sparkle    |    misty     |    crazy    |
+---+--------------+--------------+-------------+
| 0 | furry boots  | furry animal | heckin food |
| 1 | weird boots  | big animal   | furry food  |
| 2 | furry gloves | furry fence  | furry toes  |
| 3 | weird gloves | old fence    | old toes    |
| 4 | furry coat   | furry door   | furry hat   |
| 5 | weird coat   | old door     | crazy cat   |
+---+--------------+--------------+-------------+

所需的输出：

+---+--------------+--------------+-------------+---------------------------------------+
|   |   sparkle    |    misty     |    crazy    |                 furry                 |
+---+--------------+--------------+-------------+---------------------------------------+
| 0 | furry boots  | furry animal | heckin food | furry boots, furry animal             |
| 1 | weird boots  | big animal   | furry food  | furry food                            |
| 2 | furry gloves | furry fence  | furry toes  | furry gloves, furry fence, furry toes |
| 3 | weird gloves | old fence    | old toes    |                                       |
| 4 | furry coat   | furry door   | furry hat   | furry coat, furry door, furry hat     |
| 5 | weird coat   | old door     | crazy cat   |                                       |
+---+--------------+--------------+-------------+---------------------------------------+

我当前的解决方案

df['furry'] = ''
df
df.loc[df['sparkle'].str.contains('furry'), 'furry'] = df['sparkle']
df.loc[df['misty'].str.contains('furry'), 'furry'] = df['furry'] + ', ' + df['misty']
df.loc[df['crazy'].str.contains('furry'), 'furry'] = df[['furry', 'crazy']].apply(lambda x: ', '.join(x), axis=1)
df


+---+--------------+--------------+-------------+---------------------------------------+
|   |   sparkle    |    misty     |    crazy    |                 furry                 |
+---+--------------+--------------+-------------+---------------------------------------+
| 0 | furry boots  | furry animal | heckin food | furry boots, furry animal             |
| 1 | weird boots  | big animal   | furry food  | , furry food                          |
| 2 | furry gloves | furry fence  | furry toes  | furry gloves, furry fence, furry toes |
| 3 | weird gloves | old fence    | old toes    |                                       |
| 4 | furry coat   | furry door   | furry hat   | furry coat, furry door, furry hat     |
| 5 | weird coat   | old door     | crazy cat   |                                       |
+---+--------------+--------------+-------------+---------------------------------------+

这个“有效”，我可以清理后记，但是感觉很糟糕。希望在这里学习。

我正在尝试和努力的事情：

就像我在上面提到的那样，我想将其减少为读取12列，许多行以及一个单词库。我觉得我快要到了……我看过''.join（），在文档中扫描了concat（），merge（）...我只是感到困惑。

df = pd.DataFrame({'sparkle': ['furry boots', 'weird boots', 'furry gloves', 'weird gloves', 'furry coat', 'weird coat'],
      'misty': ['furry animal', 'big animal', 'furry fence', 'old fence', 'furry door', 'old door'],
      'crazy': ['heckin food', 'furry food', 'furry toes', 'old toes', 'furry hat', 'crazy cat']})
df['furry'] = ''

words = ['furry', 'old'] # added another word to demonstrate intent with real data
for key, value in df.items():
    df.loc[df[key].str.contains('|'.join(words)), 'furry'] = df['furry'] + ', ' + df[key]

df



+---+--------------+--------------+-------------+----------------------------------------------------------------------------------+
|   |   sparkle    |    misty     |    crazy    |                                      furry                                       |
+---+--------------+--------------+-------------+----------------------------------------------------------------------------------+
| 0 | furry boots  | furry animal | heckin food | , furry boots, furry animal, , furry boots, furry animal                         |
| 1 | weird boots  | big animal   | furry food  | , furry food, , furry food                                                       |
| 2 | furry gloves | furry fence  | furry toes  | , furry gloves, furry fence, furry toes, , furry gloves, furry fence, furry toes |
| 3 | weird gloves | old fence    | old toes    | , old fence, old toes, , old fence, old toes                                     |
| 4 | furry coat   | furry door   | furry hat   | , furry coat, furry door, furry hat, , furry coat, furry door, furry hat         |
| 5 | weird coat   | old door     | crazy cat   | , old door, , old door                                                           |
+---+--------------+--------------+-------------+----------------------------------------------------------------------------------+

有人有任何指示/提示吗？感谢您的阅读。

Answer 1

您可以使用apply函数

words = ['furry', 'old']
for word in words:
    df[word] = df.apply(lambda x: ', '.join([str(c) for c in x if word in str(c)]), axis=1)
df['all_combined'] = df[words].apply(lambda x:', '.join(x), axis=1)
df = df.drop(words, axis=1)

更新：您可以遍历多个单词并为每个单词创建一个新列。
Update2：同样，您可以使用apply将其合并。

解决方案2：

在回答完所有问题后，这对我来说似乎是最优雅的解决方案。

words = ['furry', 'old']
df['all_combined'] = df.apply(lambda x: ', '.join([str(c) for c in x if any([w in str(c) for w in words])]), axis=1)

有条件地将两个字符串组合成新列的优雅方法

1 个答案: