I've used pandas.read_csv to load in a file.
I've stored the file into a variable. 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.
I can't seem to find the write functionality for pandas.dataframe.
Side Note I would prefer a different library for loading in my file, but pandas provides some other different functionality which I need.
My Code:
Data = pandas.read_csv(pathFile,header=None)
doing: print Data
gives me:
0 1 2 ...
0 [2014, 8, 26, 5, 30, 0.0] 0 0.25 ...
(as you can see its a date)
Question: How to split/separate each number and save it in a new array
p.s. I'm trying to achieve the same thing the matlab method datevec()
does
答案 0 :(得分:3)
如果CSV数据看起来像
"[2014, 8, 26, 5, 30, 0.0]",0,0.25
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)
产量
0 1 2 3 4 5 6 7
0 0 0.25 2014 8 26 5 30 0
将值解析为数值。
工作原理:
dates, df = df[0], df.iloc[:, 1:]
剥离第一列,并将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
包含字符串:
In [218]: dates.iloc[0]
Out[218]: '[2014, 8, 26, 5, 30, 0.0]'
我们可以使用json.loads
将这些转换为列表:
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
我们可以使用dates
:
apply
的每一行执行此操作
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
通过上面的lambda
,返回一个系列,apply
将返回一个DataFrame,
系列的索引成为DataFrame的列索引。
现在我们可以使用pd.concat
将此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
答案 1 :(得分:0)
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'
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