For loop to compare previous values in python

Question

I have a pandas dataframe like this..

 order_id buyer_id scheduled_order  minutes   flag  
  525      232               1        13      Null   
  862      232               1        14      Null   
 1361      232               1        15      Null   
 1373      232               1        13      Null   
 1580      232               1        14      Null   
 1729      232               0        11      Null   
 1817      232               1        18      Null

I want to set a flag depending upon value of scheduled_order. If first order is scheduled order(scheduled_order =1) flag should set to 0, else it should check if minutes are greater than 12 then flag should be 1 else 2 Then, for next order if previous order is scheduled order then, flag should set to 3. If previous order is live order(scheduled_order =0) and if minutes is less than 12 then flag should set to 2. if minute is greater than 12 then flag should set to 1.

My desired output is

order_id buyer_id scheduled_order  minutes   flag  
  525      232               1        13      0   
  862      232               1        14      3   
 1361      232               1        15      3   
 1373      232               1        13      3   
 1580      232               1        14      3   
 1729      232               0        11      3   
 1817      232               1        18      2

Here is my code in python

for i in range(len(df)):
    if(df.scheduled_order[i]  == '1'):
            speed.flag[i] = '0'
    else:
        if(minutes > 12):
            df.flag[i] = '1'
        else:
            df.flag[i] = '2'

But when i becomes 1 how do I check for previous scheduled_order value?

why do have flag=2 in the last row in your output? Did you mean "... and if previous minutes is less than 12, then flag should set to 2" in your algorithm? — MaxU - stand with Ukraine
– MaxU - stand with Ukraine, Commented Feb 27, 2016 at 16:07

MaxU - stand with Ukraine · Accepted Answer · 2016-02-27 17:15:08Z

try this:

from __future__ import print_function

import pandas as pd


# create DataFrame from the CSV file
df = pd.read_csv('data.csv', delimiter=r'\s+')

# set flag to 3, for all rows where previous 'scheduled_order' == 1
# except first row
df.ix[(df.index > 0) & (df['scheduled_order'].shift(1) == 1), ['flag']] = 3

# set flag to 1, for all rows where previous 'scheduled_order' != 1
# and minutes > 12
# except first row
df.ix[(df.index > 0) & (df['scheduled_order'].shift(1) != 1) & (df['minutes'] > 12), ['flag']] = 1

# set flag to 2, for all rows where previous 'scheduled_order' != 1
# and minutes <= 12, except first row
df.ix[(df.index > 0) & (df['scheduled_order'].shift(1) != 1) & (df['minutes'] <= 12), ['flag']] = 2

# set flag for the first row ...
if df.ix[0]['scheduled_order'] == 1:
    df.ix[0, ['flag']] = 0
else:
    if df.ix[0]['minutes'] > 12:
        df.ix[0, ['flag']] = 1
    else:
        df.ix[0, ['flag']] = 2

print(df)

Output:

   order_id  buyer_id  scheduled_order  minutes flag
0       525       232                1       13    0
1       862       232                1       14    3
2      1361       232                1       15    3
3      1373       232                1       13    3
4      1580       232                1       14    3
5      1729       232                0       11    3
6      1817       232                1       18    1

PS i've followed your algorithm, that's why i have (flag == 1) for the last row. If it's not what you want, please clarify the algorithm.

IF you want to compare with "previous" minutes, then make the following replacement: df['minutes'] --> df['minutes'].shift(1), so that the output will be exactly the same as yours.

Spyke · Accepted Answer · 2016-02-27 15:29:01Z

0

You could assign scheduled_order to another temp variable and compare

answered Feb 27, 2016 at 15:29

Spyke

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Comments

Andriy Ivaneyko · Accepted Answer · 2016-02-27 15:49:13Z

0

Access prev scheduled_order as scheduled_order[i-1].

Note that you can access in that way when i > 0 otherwise your code would be buggy, because you would access last element of list.

answered Feb 27, 2016 at 15:49

Andriy Ivaneyko

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Comments

Gustavo Bezerra · Accepted Answer · 2016-02-27 15:51:31Z

0

Add a column with the previous scheduled order value:

df['prev_scheduled_order'] = df.scheduled_order.shift(1)

answered Feb 27, 2016 at 15:51

Gustavo Bezerra

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