在 Python / Pandas 中计算两行之间的差异

Question

In python, how can I reference previous row and calculate something against it? Specifically, I am working with dataframes in pandas - I have a data frame full of stock price information that looks like this:

           Date   Close  Adj Close
251  2011-01-03  147.48     143.25
250  2011-01-04  147.64     143.41
249  2011-01-05  147.05     142.83
248  2011-01-06  148.66     144.40
247  2011-01-07  147.93     143.69

Here is how I created this dataframe:

import pandas

url = 'http://ichart.finance.yahoo.com/table.csv?s=IBM&a=00&b=1&c=2011&d=11&e=31&f=2011&g=d&ignore=.csv'
data = data = pandas.read_csv(url)

## now I sorted the data frame ascending by date 
data = data.sort(columns='Date')

Starting with row number 2, or in this case, I guess it's 250 (PS - is that the index?), I want to calculate the difference between 2011-01-03 and 2011-01-04, for every entry in this dataframe. I believe the appropriate way is to write a function that takes the current row, then figures out the previous row, and calculates the difference between them, the use the pandas apply function to update the dataframe with the value.

Is that the right approach? If so, should I be using the index to determine the difference? (note - I'm still in python beginner mode, so index may not be the right term, nor even the correct way to implement this)

Answer 1

I think you want to do something like this:

In [26]: data
Out[26]: 
           Date   Close  Adj Close
251  2011-01-03  147.48     143.25
250  2011-01-04  147.64     143.41
249  2011-01-05  147.05     142.83
248  2011-01-06  148.66     144.40
247  2011-01-07  147.93     143.69

In [27]: data.set_index('Date').diff()
Out[27]: 
            Close  Adj Close
Date                        
2011-01-03    NaN        NaN
2011-01-04   0.16       0.16
2011-01-05  -0.59      -0.58
2011-01-06   1.61       1.57
2011-01-07  -0.73      -0.71

Answer 2

To calculate difference of one column. Here is what you can do.

df=
      A      B
0     10     56
1     45     48
2     26     48
3     32     65

We want to compute row difference in A only and want to consider the rows which are less than 15.

df['A_dif'] = df['A'].diff()
df=
          A      B      A_dif
    0     10     56      Nan
    1     45     48      35
    2     26     48      19
    3     32     65      6
df = df[df['A_dif']<15]

df=
          A      B      A_dif
    0     10     56      Nan
    3     32     65      6

Answer 3

I don't know pandas, and I'm pretty sure it has something specific for this; however, I'll give you the pure-Python solution, that might be of some help even if you need to use pandas:

import csv
import urllib

# This basically retrieves the CSV files and loads it in a list, converting
# All numeric values to floats
url='http://ichart.finance.yahoo.com/table.csv?s=IBM&a=00&b=1&c=2011&d=11&e=31&f=2011&g=d&ignore=.csv'
reader = csv.reader(urllib.urlopen(url), delimiter=',')
# We sort the output list so the records are ordered by date
cleaned = sorted([[r[0]] + map(float, r[1:]) for r in list(reader)[1:]])

for i, row in enumerate(cleaned):  # enumerate() yields two-tuples: (<id>, <item>)
    # The try..except here is to skip the IndexError for line 0
    try:
        # This will calculate difference of each numeric field with the same field
        # in the row before this one
        print row[0], [(row[j] - cleaned[i-1][j]) for j in range(1, 7)]
    except IndexError:
        pass