[英]Python pandas.read_csv split column into multiple new columns using comma to separate
I've used pandas.read_csv to load in a file. 我使用pandas.read_csv加载文件。
I've stored the file into a variable. 我已将文件存储到变量中。 The first column is a series of numbers separated by a comma (,) I want to split these numbers, and put each number to a new column.
第一列是一系列数字,中间用逗号(,)分隔,我想将这些数字分割,然后将每个数字放到新列中。
I can't seem to find the write functionality for pandas.dataframe. 我似乎找不到pandas.dataframe的写入功能。
Side Note I would prefer a different library for loading in my file, but pandas provides some other different functionality which I need. 旁注我希望使用其他库来加载文件,但是pandas提供了我需要的其他一些功能。
My Code: 我的代码:
Data = pandas.read_csv(pathFile,header=None)
doing: print Data
gives me: 这样做:
print Data
给我:
0 1 2 ...
0 [2014, 8, 26, 5, 30, 0.0] 0 0.25 ...
(as you can see its a date) (如您所见,它是一个日期)
Question: How to split/separate each number and save it in a new array 问题:如何分割/分隔每个数字并将其保存在新数组中
ps I'm trying to achieve the same thing the matlab method datevec()
does ps我正在尝试实现matlab方法
datevec()
所做的相同操作
If the CSV data looks like 如果CSV数据看起来像
"[2014, 8, 26, 5, 30, 0.0]",0,0.25
then 然后
import pandas as pd
import json
df = pd.read_csv('data', header=None)
dates, df = df[0], df.iloc[:, 1:]
df = pd.concat([df, dates.apply(lambda x: pd.Series(json.loads(x)))], axis=1,
ignore_index=True)
print(df)
yields 产量
0 1 2 3 4 5 6 7
0 0 0.25 2014 8 26 5 30 0
with the values parsed as numeric values. 将值解析为数字值。
How it works: 这个怎么运作:
dates, df = df[0], df.iloc[:, 1:]
peels off the first column, and reassigns df
to the rest of the DataFrame: 剥离第一列,然后将
df
重新分配给DataFrame的其余部分:
In [217]: dates
Out[217]:
0 [2014, 8, 26, 5, 30, 0.0]
Name: 0, dtype: object
dates
contains strings: dates
包含字符串:
In [218]: dates.iloc[0]
Out[218]: '[2014, 8, 26, 5, 30, 0.0]'
We can convert these to a list using json.loads
: 我们可以使用
json.loads
将它们转换为列表:
In [219]: import json
In [220]: json.loads(dates.iloc[0])
Out[220]: [2014, 8, 26, 5, 30, 0.0]
In [221]: type(json.loads(dates.iloc[0]))
Out[221]: list
We can do this for each row of dates
by using apply
: 我们可以使用
apply
对每一行dates
执行此操作:
In [222]: dates.apply(lambda x: pd.Series(json.loads(x)))
Out[222]:
0 1 2 3 4 5
0 2014 8 26 5 30 0
By making lambda
, above, return a Series, apply
will return a DataFrame, with the index of the Series becoming the column index of the DataFrame. 通过在上面创建
lambda
,返回一个Series, apply
将返回一个DataFrame,Series的索引成为该DataFrame的列索引。
Now we can use pd.concat
to concatenate this DataFrame with df
: 现在我们可以使用
pd.concat
将这个DataFrame与df
连接起来:
In [228]: df = pd.concat([df, dates.apply(lambda x: pd.Series(json.loads(x)))], axis=1, ignore_index=True)
In [229]: df
Out[229]:
0 1 2 3 4 5 6 7
0 0 0.25 2014 8 26 5 30 0
In [230]: df.dtypes
Out[230]:
0 int64
1 float64
2 float64
3 float64
4 float64
5 float64
6 float64
7 float64
dtype: object
How about 怎么样
df
# datestr
#0 2014, 8, 26, 5, 30, 0.0
#1 2014, 8, 26, 5, 30, 0.0
#2 2014, 8, 26, 5, 30, 0.0
#3 2014, 8, 26, 5, 30, 0.0
#4 2014, 8, 26, 5, 30, 0.0
# each entry is a string
df.datestr[0]
#'2014, 8, 26, 5, 30, 0.0'
Then 然后
date_order = ('year', 'month','day','hour','minute','sec') # order matters here, should match the datestr column
for i,col in enumerate( date_order):
df[col] = df.datestr.map( lambda x: x.split(',')[i].strip() )
#df
# datestr year month day hour minute sec
#0 2014, 8, 26, 5, 30, 0.0 2014 8 26 5 30 0.0
#1 2014, 8, 26, 5, 30, 0.0 2014 8 26 5 30 0.0
#2 2014, 8, 26, 5, 30, 0.0 2014 8 26 5 30 0.0
#3 2014, 8, 26, 5, 30, 0.0 2014 8 26 5 30 0.0
#4 2014, 8, 26, 5, 30, 0.0 2014 8 26 5 30 0.0
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