我有一个只有一列的表。我想将我编写的函数应用于该系列的其他每一行。但是,当我这样做时,我会得到一个错误!
The table looks like this: And I want to get this:
names names
bank account bank account|bank|account
1256864 1256864
bank share bank share|bank|share
42,566 42,566
bank currency bank currency|bank|currency
Dollar Dollar
batch number batch number|batch|number
001444 001444
... ...
这是我编写的代码:
import pandas as pd
import re
df = pd.read_table('list_a.tsv')
def sep_rows (text):
sperated = '|'.join(re.split(r'\s+', text))
return text+'|'+sperated
# this applies the function to ALL rows!
print(df['names'].apply(sep_rows))
# I tried to choose every other row
a = df.iloc[::2].apply(sep_rows)
print(a) # But I gen an error!
我明白了:
TypeError: expected string or bytes-like object
答案 0 :(得分:3)
您的方法(使用re
和apply
)过于复杂且缓慢。以下表达式使用本地Pandas向量化,效率更高(运行速度快约4倍)。
evens = df['names'].iloc[::2]
evens[:] = evens + '|' + evens.str.replace('\s+', '|')
# names
#0 bank account|bank|account
#1 1256864
#2 bank share|bank|share
#3 42,566
答案 1 :(得分:1)
按顺序处理文本,然后您的功能应该可以工作:
def sep_rows(text):
separated = text.str.replace(r"\s+", "|")
return text + "|" + separated
df.iloc[::2].apply(sep_rows)
names
0 bank account|bank|account
2 bank share|bank|share
4 bank currency|bank|currency
6 batch number|batch|number
获得结果的另一种方法是list comprehension
:
import re
df['new_column'] = ["|".join((text, re.sub(r"\s+", "|", text)))
if num%2 ==0 else text
for num, text in enumerate(df.names)
]
df
names new_column
0 bank account bank account|bank|account
1 1256864 1256864
2 bank share bank share|bank|share
3 42,566 42,566
4 bank currency bank currency|bank|currency
5 Dollar Dollar
6 batch number batch number|batch|number
7 001444 001444