Question

目标是根据现有列的数据框的行级信息的返回值创建新列。

df = pd.DataFrame({"A": [The quick brown fox jumps over the lazy dog,Glib jocks quiz nymph to vex dwarf], "B": [10, 20]})

                                             A   B
0  The quick brown fox jumps over the lazy dog  10
1           Glib jocks quiz nymph to vex dwarf  20

存在方法：

def returnTopic(model, query, numberOftopics): 
    # strip out topics per query/row and return topics that are relevant to a query/row
    return topicDict`

topicDict包含{'x': ['fox','brown'], 'y':['jumps','over','the']}

我想从字典中的这些返回元素创建两个新列。

                                             A   B  x               y
0  The quick brown fox jumps over the lazy dog  10 ['fox','brown'] ['jumps','over','the']
1           Glib jocks quiz nymph to vex dwarf  20

这是我的尝试：

df['x'] = df.apply(lambda x: returnTopic(tmodel['x'], x['A'], 2))

Answer 1

您可以从记录中创建新的DataFrame，并将其连接到旧的DataFrame。

像这样：

import pandas as pd
df = pd.DataFrame({"A": ['The quick brown fox jumps over the lazy dog',
                         'Glib jocks quiz nymph to vex dwarf'], "B": [10, 20]})

def f(text, something_else):
    return {'x':len(text), 'y': text.count(' ')}

new_df = pd.concat([df, pd.DataFrame.from_records(df['A'].apply(lambda x: f(x, 0)))], axis=1)
print(new_df)

它会返回

                                             A   B   x  y
0  The quick brown fox jumps over the lazy dog  10  43  8
1           Glib jocks quiz nymph to vex dwarf  20  34  6

Answer 2

让你的函数返回一个pd.Series对象：

def foo(x): 
    ...
    return pd.Series(topicDict)

现在，沿第一个轴调用apply：

v = df.apply(foo, 1)
v
              x                   y
0  [fox, brown]  [jumps, over, the]
1  [fox, brown]  [jumps, over, the]

使用pd.concat将结果与原始内容连接起来。

pd.concat([df, v], 1)

                                             A   B             x  \
0  The quick brown fox jumps over the lazy dog  10  [fox, brown]   
1           Glib jocks quiz nymph to vex dwarf  20  [fox, brown]   

                    y  
0  [jumps, over, the]  
1  [jumps, over, the]

Pandas通过应用返回字典的函数创建一个新列

2 个答案: