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在熊猫数据框中将Unix时间转换为可读日期

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问题：在熊猫数据框中将Unix时间转换为可读日期
回答 0
回答 1
回答 2
回答 3

问题：在熊猫数据框中将Unix时间转换为可读日期

我有一个带有unix时间和价格的数据框。我想转换索引列，以便以人类可读的日期显示它。

因此，例如，我在index列中有dateas 1349633705，但我希望它显示为10/07/2012（或至少10/07/2012 18:15）。

在某些情况下，这是我正在使用的代码以及我已经尝试过的代码：

import json
import urllib2
from datetime import datetime
response = urllib2.urlopen('http://blockchain.info/charts/market-price?&format=json')
data = json.load(response)   
df = DataFrame(data['values'])
df.columns = ["date","price"]
#convert dates 
df.date = df.date.apply(lambda d: datetime.strptime(d, "%Y-%m-%d"))
df.index = df.date

如您所见，我在df.date = df.date.apply(lambda d: datetime.strptime(d, "%Y-%m-%d"))这里使用的 是无效的，因为我使用的是整数而不是字符串。我认为我需要使用，datetime.date.fromtimestamp但我不确定如何将其应用于整个df.date。

谢谢。

点击查看英文原文

I have a dataframe with unix times and prices in it. I want to convert the index column so that it shows in human readable dates.

So for instance I have date as 1349633705 in the index column but I’d want it to show as 10/07/2012 (or at least 10/07/2012 18:15).

For some context, here is the code I’m working with and what I’ve tried already:

import json
import urllib2
from datetime import datetime
response = urllib2.urlopen('http://blockchain.info/charts/market-price?&format=json')
data = json.load(response)   
df = DataFrame(data['values'])
df.columns = ["date","price"]
#convert dates 
df.date = df.date.apply(lambda d: datetime.strptime(d, "%Y-%m-%d"))
df.index = df.date

As you can see I’m using df.date = df.date.apply(lambda d: datetime.strptime(d, "%Y-%m-%d")) here which doesn’t work since I’m working with integers, not strings. I think I need to use datetime.date.fromtimestamp but I’m not quite sure how to apply this to the whole of df.date.

Thanks.

回答 0

距纪元似乎只有几秒钟。

In [20]: df = DataFrame(data['values'])

In [21]: df.columns = ["date","price"]

In [22]: df
Out[22]: 
<class 'pandas.core.frame.DataFrame'>
Int64Index: 358 entries, 0 to 357
Data columns (total 2 columns):
date     358  non-null values
price    358  non-null values
dtypes: float64(1), int64(1)

In [23]: df.head()
Out[23]: 
         date  price
0  1349720105  12.08
1  1349806505  12.35
2  1349892905  12.15
3  1349979305  12.19
4  1350065705  12.15
In [25]: df['date'] = pd.to_datetime(df['date'],unit='s')

In [26]: df.head()
Out[26]: 
                 date  price
0 2012-10-08 18:15:05  12.08
1 2012-10-09 18:15:05  12.35
2 2012-10-10 18:15:05  12.15
3 2012-10-11 18:15:05  12.19
4 2012-10-12 18:15:05  12.15

In [27]: df.dtypes
Out[27]: 
date     datetime64[ns]
price           float64
dtype: object
点击查看英文原文

These appear to be seconds since epoch.

In [20]: df = DataFrame(data['values'])

In [21]: df.columns = ["date","price"]

In [22]: df
Out[22]: 
<class 'pandas.core.frame.DataFrame'>
Int64Index: 358 entries, 0 to 357
Data columns (total 2 columns):
date     358  non-null values
price    358  non-null values
dtypes: float64(1), int64(1)

In [23]: df.head()
Out[23]: 
         date  price
0  1349720105  12.08
1  1349806505  12.35
2  1349892905  12.15
3  1349979305  12.19
4  1350065705  12.15
In [25]: df['date'] = pd.to_datetime(df['date'],unit='s')

In [26]: df.head()
Out[26]: 
                 date  price
0 2012-10-08 18:15:05  12.08
1 2012-10-09 18:15:05  12.35
2 2012-10-10 18:15:05  12.15
3 2012-10-11 18:15:05  12.19
4 2012-10-12 18:15:05  12.15

In [27]: df.dtypes
Out[27]: 
date     datetime64[ns]
price           float64
dtype: object

回答 1

如果您尝试使用： 

df[DATE_FIELD]=(pd.to_datetime(df[DATE_FIELD],***unit='s'***))

并收到一个错误：

“ pandas.tslib.OutOfBoundsDatetime：无法转换单位为’s’的输入”

这表示DATE_FIELD未指定秒。

就我而言，是毫秒- EPOCH time。

转换工作如下：

df[DATE_FIELD]=(pd.to_datetime(df[DATE_FIELD],unit='ms'))
点击查看英文原文

If you try using: 

df[DATE_FIELD]=(pd.to_datetime(df[DATE_FIELD],***unit='s'***))

and receive an error :

“pandas.tslib.OutOfBoundsDatetime: cannot convert input with unit ‘s'”

This means the DATE_FIELD is not specified in seconds.

In my case, it was milli seconds – EPOCH time.

The conversion worked using below:

df[DATE_FIELD]=(pd.to_datetime(df[DATE_FIELD],unit='ms'))

回答 2

假设我们导入了pandas as pd并且df是我们的数据框

pd.to_datetime(df['date'], unit='s')

为我工作。

点击查看英文原文

Assuming we imported pandas as pd and df is our dataframe

pd.to_datetime(df['date'], unit='s')

works for me.

回答 3

或者，通过更改上面的代码行：

# df.date = df.date.apply(lambda d: datetime.strptime(d, "%Y-%m-%d"))
df.date = df.date.apply(lambda d: datetime.datetime.fromtimestamp(int(d)).strftime('%Y-%m-%d'))

它也应该起作用。

点击查看英文原文

Alternatively, by changing a line of the above code:

# df.date = df.date.apply(lambda d: datetime.strptime(d, "%Y-%m-%d"))
df.date = df.date.apply(lambda d: datetime.datetime.fromtimestamp(int(d)).strftime('%Y-%m-%d'))

It should also work.