我有一个csv文件,其中有几列我首先用冒号(;)分隔。但是,一列由管道分隔我想分界这个专栏并创建新专栏。
输入:
Column 1 Column 2 Column 3
1 2 3|4|5
6 7 6|7|8
10 11 12|13|14
期望的输出:
Column 1 Column 2 ID Age Height
1 2 3 4 5
6 7 6 7 8
10 11 12 13 14
我的代码到目前为止第一次划分;然后转换为DF(这是我想要的结束格式)</ p>
delimit = list(csv.reader(open('test.csv', 'rt'), delimiter=';'))
df = pd.DataFrame(delimit)
答案 0 :(得分:3)
你没有准确显示数据的样子(你说它用分号分隔,但你的例子没有),但如果它看起来像
Column 1;Column 2;Column 3
1;2;3|4|5
6;7;6|7|8
10;11;12|13|14
您可以执行类似
的操作>>> df = pd.read_csv("test.csv", sep="[;|]", engine='python', skiprows=1,
names=["Column 1", "Column 2", "ID", "Age", "Height"])
>>> df
Column 1 Column 2 ID Age Height
0 1 2 3 4 5
1 6 7 6 7 8
2 10 11 12 13 14
这可以通过使用正则表达式分隔符来表示&#34; ;
或|
&#34;并手动强制列名称。
或者,您可以通过几个步骤完成此操作:
>>> df = pd.read_csv("test.csv", sep=";")
>>> df
Column 1 Column 2 Column 3
0 1 2 3|4|5
1 6 7 6|7|8
2 10 11 12|13|14
>>> c3 = df.pop("Column 3").str.split("|", expand=True)
>>> c3.columns = ["ID", "Age", "Height"]
>>> df.join(c3)
Column 1 Column 2 ID Age Height
0 1 2 3 4 5
1 6 7 6 7 8
2 10 11 12 13 14
答案 1 :(得分:0)
delimit = list(csv.reader(open('test.csv', 'rt'), delimiter=';'))
for row in delimit:
piped = row.pop()
row.extend(piped.split('|'))
df = pd.DataFrame(delimit)
delimit
最终看起来像:
[
['1', '2', '3', '4', '5'],
['6', '7', '6', '7', '8'],
['10', '11', '12', '13', '14'],
]
答案 2 :(得分:0)
使用csv lib和str.replace实际上要快得多:
import csv
with open("test.txt") as f:
next(f)
# itertools.imap python2
df = pd.DataFrame.from_records(csv.reader(map(lambda x: x.rstrip().replace("|", ";"), f), delimiter=";"),
columns=["Column 1", "Column 2", "ID", "Age", "Height"]).astype(int)
一些时间:
In [35]: %%timeit
pd.read_csv("test.txt", sep="[;|]", engine='python', skiprows=1,
names=["Column 1", "Column 2", "ID", "Age", "Height"])
....:
100 loops, best of 3: 14.7 ms per loop
In [36]: %%timeit
with open("test.txt") as f:
next(f)
df = pd.DataFrame.from_records(csv.reader(map(lambda x: x.rstrip().replace("|", ";"), f),delimiter=";"),
columns=["Column 1", "Column 2", "ID", "Age", "Height"]).astype(int)
....:
100 loops, best of 3: 6.05 ms per loop
你可以只是str.split:
with open("test.txt") as f:
next(f)
df = pd.DataFrame.from_records(map(lambda x: x.rstrip().replace("|", ";").split(";"), f),
columns=["Column 1", "Column 2", "ID", "Age", "Height"])
答案 3 :(得分:0)
为自己找出解决方案:
df = pd.DataFrame(delimit)
s = df['Column 3'].apply(lambda x: pd.Series(x.split('|')))
frame = pd.DataFrame(s)
frame.rename(columns={0: 'ID',1:'Height',2:'Age'}, inplace=True)
result = pd.concat([df, frame], axis=1)