Question

假设我有一个如下数据框：

import pandas as pd

df1 = pd.DataFrame({
    'A' : ['foo ', 'b,ar', 'fo...o', 'bar', 'foo', 'bar', 'foo', 'foo'],
    'B' : ['one', 'one', 'two', 'three','two', 'two', 'one', 'three'],
})

我想创建一个新数据框，df2，它是df1中“ A”和“ B”列的串联形式，其中每个数据都大写。这是一个玩具示例，在我的用例中，我可能还拥有比列“ A”和“ B”更多的列，因此我想使列的列表可变（即列名）。列可能会有所不同）。

def tokenize(s):
    # replaces comma with space; removes non-alphanumeric chars; etc.
    return re.sub('[^0-9a-zA-Z\s]+', '', re.sub('[,]+', ' ', s)).lower().split()

df2 = pd.DataFrame() # create a new dataframe; not sure if I'm doing this right
cols_to_concat = ['A','B'] # there can be more than two columns in this list
for col in cols_to_concat:
    df2 = df1[col].apply(tokenize).apply(str.upper)
print(df2)
# here, I'd like the df2 to have just ONE column whose rows are 'FOOONE', 'BARONE', 'FOOTWO', 'BARTHREE','FOOTWO', 'BARTWO','FOOONE','FOOTHREE',...

Answer 1

简短版

list_o_cols = ['A', 'B']

df1[list_o_cols].sum(1).str.upper()

0      FOOONE
1      BARONE
2      FOOTWO
3    BARTHREE
4      FOOTWO
5      BARTWO
6      FOOONE
7    FOOTHREE
dtype: object

df2 = df1[list_o_cols].sum(1).str.upper().str.replace('O', '').to_frame('col_name')
df2

   col_name
0       FNE
1     BARNE
2       FTW
3  BARTHREE
4       FTW
5     BARTW
6       FNE
7    FTHREE

Answer 2

ConcatCol = ['A', 'B']

df2 = pd.DataFrame(df1[ConcatCol].apply(lambda x: ''.join(x.str.upper()), axis=1), columns=['Col'])

根据您的评论，您可以在lambda函数之后应用函数，如果要进行串联然后应用函数：

# your function
def tokenize(s):
    # replaces comma with space; removes non-alphanumeric chars; etc.
    return re.sub('[^0-9a-zA-Z\s]+', '', re.sub('[,]+', ' ', s)).lower().split()

ConcatCol = ['A', 'B']

df2 = pd.DataFrame(df1[ConcatCol].apply(lambda x:  ''.join(x), axis=1).apply(tokenize), columns=['Col'])

       Col
0   [foo, one]
1   [b, arone]
2   [footwo]
3   [barthree]
4   [footwo]
5   [bartwo]
6   [fooone]
7   [foothree]

要先应用您的函数，然后再使用concat，答案会稍有不同，因为您的函数使用split创建列表。因此，最终，您将使用sum将列表组合在一起：

def tokenize(s):
    # replaces comma with space; removes non-alphanumeric chars; etc.
    return re.sub('[^0-9a-zA-Z\s]+', '', re.sub('[,]+', ' ', s)).lower().split()

ConcatCol = ['A', 'B']

df2 = pd.DataFrame(df1[ConcatCol].apply(lambda x: (x.apply(tokenize))).sum(axis=1), columns=['Col'])

       Col
0   [foo, one]
1   [b, ar, one]
2   [foo, two]
3   [bar, three]
4   [foo, two]
5   [bar, two]
6   [foo, one]
7   [foo, three]

应用转换并连接现有数据框中的多列以在Pandas中形成新数据框

2 个答案:

简短版