Calculate Price Change (in last 2 mins) and Price Range (before 2 mins)
0
I have a dataframe as mentioned below:
row_no,last_price,time
01,110.50,10:09
02,111.60,10:09
03,111.50,10:09
04,112.00,10:09
05,112.00,10:10
06,112.60,10:10
07,112.50,10:10
08,113.10,10:10
09,114.30,10:11
10,114.50,10:11
11,115.70,10:11
12,116.50,10:12
13,116.30,10:12
14,116.20,10:12
15,116.50,10:13
16,117.80,10:13
17,117.90,10:13
18,117.50,10:14
19,118.70,10:14
20,118.90,10:14
21,118.30,10:14
22,118.50,10:15
23,119.60,10:15
24,119.50,10:15
25,119.80,10:15
Row_No Column is added for understanding purpose.
Example for Last Row = 25: Values should be calculated as below:
- Need to Calculate "Price_Change_in_last_2mins" in new Column i.e.
(row 25 last_price - row 15 last_price)*100/ (row 15 last_price) - Need to Calculate "Price_Range_before_2mins" (period 2 min) in new Column i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
I want the result like:
row_no,last_price,time,Price_Change_in_last_2mins,Price_Range_before_2mins
01,110.50,10:09,NaN,NaN
02,111.60,10:09,NaN,NaN
03,111.50,10:09,NaN,NaN
04,112.00,10:09,NaN,NaN
05,112.00,10:10,NaN,NaN
06,112.60,10:10,NaN,NaN
07,112.50,10:10,NaN,NaN
08,113.10,10:10,NaN,NaN
09,114.30,10:11,NaN,NaN
10,114.50,10:11,NaN,NaN
11,115.70,10:11,NaN,NaN
12,116.50,10:12,NaN,NaN
13,116.30,10:12,NaN,NaN
14,116.20,10:12,NaN,NaN
15,116.50,10:13,NaN,NaN
16,117.80,10:13,NaN,NaN
17,117.90,10:13,,NaN,NaN
18,117.50,10:14,0.85,4.49
19,118.70,10:14,1.88,4.49
20,118.90,10:14,2.06,4.49
21,118.30,10:14,1.55,4.49
22,118.50,10:15,1.72,3.86
23,119.60,10:15,2.66,3.86
24,119.50,10:15,2.57,3.86
25,119.80,10:15,2.83,3.86
python python-3.x pandas dataframe list-comprehension
asked Nov 23 '18 at 10:48
PravatPravat
8510
|
show 5 more comments
0
I have a dataframe as mentioned below:
row_no,last_price,time
01,110.50,10:09
02,111.60,10:09
03,111.50,10:09
04,112.00,10:09
05,112.00,10:10
06,112.60,10:10
07,112.50,10:10
08,113.10,10:10
09,114.30,10:11
10,114.50,10:11
11,115.70,10:11
12,116.50,10:12
13,116.30,10:12
14,116.20,10:12
15,116.50,10:13
16,117.80,10:13
17,117.90,10:13
18,117.50,10:14
19,118.70,10:14
20,118.90,10:14
21,118.30,10:14
22,118.50,10:15
23,119.60,10:15
24,119.50,10:15
25,119.80,10:15
Row_No Column is added for understanding purpose.
Example for Last Row = 25: Values should be calculated as below:
- Need to Calculate "Price_Change_in_last_2mins" in new Column i.e.
(row 25 last_price - row 15 last_price)*100/ (row 15 last_price) - Need to Calculate "Price_Range_before_2mins" (period 2 min) in new Column i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
I want the result like:
row_no,last_price,time,Price_Change_in_last_2mins,Price_Range_before_2mins
01,110.50,10:09,NaN,NaN
02,111.60,10:09,NaN,NaN
03,111.50,10:09,NaN,NaN
04,112.00,10:09,NaN,NaN
05,112.00,10:10,NaN,NaN
06,112.60,10:10,NaN,NaN
07,112.50,10:10,NaN,NaN
08,113.10,10:10,NaN,NaN
09,114.30,10:11,NaN,NaN
10,114.50,10:11,NaN,NaN
11,115.70,10:11,NaN,NaN
12,116.50,10:12,NaN,NaN
13,116.30,10:12,NaN,NaN
14,116.20,10:12,NaN,NaN
15,116.50,10:13,NaN,NaN
16,117.80,10:13,NaN,NaN
17,117.90,10:13,,NaN,NaN
18,117.50,10:14,0.85,4.49
19,118.70,10:14,1.88,4.49
20,118.90,10:14,2.06,4.49
21,118.30,10:14,1.55,4.49
22,118.50,10:15,1.72,3.86
23,119.60,10:15,2.66,3.86
24,119.50,10:15,2.57,3.86
25,119.80,10:15,2.83,3.86
python python-3.x pandas dataframe list-comprehension
asked Nov 23 '18 at 10:48
PravatPravat
8510
2
Have you tried to solve these yourself? Can you share with us some attempts at solving it?
– yatu
Nov 23 '18 at 10:50
I think it will help you stackoverflow.com/questions/12376863/… use a.apply(command)
– Jony Kim
Nov 23 '18 at 10:56
Sir, I am new to python and unable to handle data with calculating Time Difference.
– Pravat
Nov 23 '18 at 11:06
I see what you mean. You might want to specify that the example of rows 15 and 25 that you provide is the calculation for row 25...
– yatu
Nov 23 '18 at 11:07
@Pravat, how do you obtain 0.03 in row 24?
– yatu
Nov 23 '18 at 11:49
|
show 5 more comments
0
0
0
1
I have a dataframe as mentioned below:
row_no,last_price,time
01,110.50,10:09
02,111.60,10:09
03,111.50,10:09
04,112.00,10:09
05,112.00,10:10
06,112.60,10:10
07,112.50,10:10
08,113.10,10:10
09,114.30,10:11
10,114.50,10:11
11,115.70,10:11
12,116.50,10:12
13,116.30,10:12
14,116.20,10:12
15,116.50,10:13
16,117.80,10:13
17,117.90,10:13
18,117.50,10:14
19,118.70,10:14
20,118.90,10:14
21,118.30,10:14
22,118.50,10:15
23,119.60,10:15
24,119.50,10:15
25,119.80,10:15
Row_No Column is added for understanding purpose.
Example for Last Row = 25: Values should be calculated as below:
- Need to Calculate "Price_Change_in_last_2mins" in new Column i.e.
(row 25 last_price - row 15 last_price)*100/ (row 15 last_price) - Need to Calculate "Price_Range_before_2mins" (period 2 min) in new Column i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
I want the result like:
row_no,last_price,time,Price_Change_in_last_2mins,Price_Range_before_2mins
01,110.50,10:09,NaN,NaN
02,111.60,10:09,NaN,NaN
03,111.50,10:09,NaN,NaN
04,112.00,10:09,NaN,NaN
05,112.00,10:10,NaN,NaN
06,112.60,10:10,NaN,NaN
07,112.50,10:10,NaN,NaN
08,113.10,10:10,NaN,NaN
09,114.30,10:11,NaN,NaN
10,114.50,10:11,NaN,NaN
11,115.70,10:11,NaN,NaN
12,116.50,10:12,NaN,NaN
13,116.30,10:12,NaN,NaN
14,116.20,10:12,NaN,NaN
15,116.50,10:13,NaN,NaN
16,117.80,10:13,NaN,NaN
17,117.90,10:13,,NaN,NaN
18,117.50,10:14,0.85,4.49
19,118.70,10:14,1.88,4.49
20,118.90,10:14,2.06,4.49
21,118.30,10:14,1.55,4.49
22,118.50,10:15,1.72,3.86
23,119.60,10:15,2.66,3.86
24,119.50,10:15,2.57,3.86
25,119.80,10:15,2.83,3.86
python python-3.x pandas dataframe list-comprehension
asked Nov 23 '18 at 10:48
PravatPravat
8510
I have a dataframe as mentioned below:
row_no,last_price,time
01,110.50,10:09
02,111.60,10:09
03,111.50,10:09
04,112.00,10:09
05,112.00,10:10
06,112.60,10:10
07,112.50,10:10
08,113.10,10:10
09,114.30,10:11
10,114.50,10:11
11,115.70,10:11
12,116.50,10:12
13,116.30,10:12
14,116.20,10:12
15,116.50,10:13
16,117.80,10:13
17,117.90,10:13
18,117.50,10:14
19,118.70,10:14
20,118.90,10:14
21,118.30,10:14
22,118.50,10:15
23,119.60,10:15
24,119.50,10:15
25,119.80,10:15
Row_No Column is added for understanding purpose.
Example for Last Row = 25: Values should be calculated as below:
- Need to Calculate "Price_Change_in_last_2mins" in new Column i.e.
(row 25 last_price - row 15 last_price)*100/ (row 15 last_price) - Need to Calculate "Price_Range_before_2mins" (period 2 min) in new Column i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
I want the result like:
row_no,last_price,time,Price_Change_in_last_2mins,Price_Range_before_2mins
01,110.50,10:09,NaN,NaN
02,111.60,10:09,NaN,NaN
03,111.50,10:09,NaN,NaN
04,112.00,10:09,NaN,NaN
05,112.00,10:10,NaN,NaN
06,112.60,10:10,NaN,NaN
07,112.50,10:10,NaN,NaN
08,113.10,10:10,NaN,NaN
09,114.30,10:11,NaN,NaN
10,114.50,10:11,NaN,NaN
11,115.70,10:11,NaN,NaN
12,116.50,10:12,NaN,NaN
13,116.30,10:12,NaN,NaN
14,116.20,10:12,NaN,NaN
15,116.50,10:13,NaN,NaN
16,117.80,10:13,NaN,NaN
17,117.90,10:13,,NaN,NaN
18,117.50,10:14,0.85,4.49
19,118.70,10:14,1.88,4.49
20,118.90,10:14,2.06,4.49
21,118.30,10:14,1.55,4.49
22,118.50,10:15,1.72,3.86
23,119.60,10:15,2.66,3.86
24,119.50,10:15,2.57,3.86
25,119.80,10:15,2.83,3.86
python python-3.x pandas dataframe list-comprehension
python python-3.x pandas dataframe list-comprehension
asked Nov 23 '18 at 10:48
PravatPravat
8510
asked Nov 23 '18 at 10:48
PravatPravat
8510
edited Nov 23 '18 at 12:02
Pravat
asked Nov 23 '18 at 10:48
PravatPravat
8510
asked Nov 23 '18 at 10:48
PravatPravat
8510
asked Nov 23 '18 at 10:48
PravatPravat
8510
8510
2
Have you tried to solve these yourself? Can you share with us some attempts at solving it?
– yatu
Nov 23 '18 at 10:50
I think it will help you stackoverflow.com/questions/12376863/… use a.apply(command)
– Jony Kim
Nov 23 '18 at 10:56
Sir, I am new to python and unable to handle data with calculating Time Difference.
– Pravat
Nov 23 '18 at 11:06
I see what you mean. You might want to specify that the example of rows 15 and 25 that you provide is the calculation for row 25...
– yatu
Nov 23 '18 at 11:07
@Pravat, how do you obtain 0.03 in row 24?
– yatu
Nov 23 '18 at 11:49
|
show 5 more comments
2
Have you tried to solve these yourself? Can you share with us some attempts at solving it?
– yatu
Nov 23 '18 at 10:50
I think it will help you stackoverflow.com/questions/12376863/… use a.apply(command)
– Jony Kim
Nov 23 '18 at 10:56
Sir, I am new to python and unable to handle data with calculating Time Difference.
– Pravat
Nov 23 '18 at 11:06
I see what you mean. You might want to specify that the example of rows 15 and 25 that you provide is the calculation for row 25...
– yatu
Nov 23 '18 at 11:07
@Pravat, how do you obtain 0.03 in row 24?
– yatu
Nov 23 '18 at 11:49
2
2
Have you tried to solve these yourself? Can you share with us some attempts at solving it?
– yatu
Nov 23 '18 at 10:50
Have you tried to solve these yourself? Can you share with us some attempts at solving it?
– yatu
Nov 23 '18 at 10:50
I think it will help you stackoverflow.com/questions/12376863/… use a.apply(command)
– Jony Kim
Nov 23 '18 at 10:56
I think it will help you stackoverflow.com/questions/12376863/… use a.apply(command)
– Jony Kim
Nov 23 '18 at 10:56
Sir, I am new to python and unable to handle data with calculating Time Difference.
– Pravat
Nov 23 '18 at 11:06
Sir, I am new to python and unable to handle data with calculating Time Difference.
– Pravat
Nov 23 '18 at 11:06
I see what you mean. You might want to specify that the example of rows 15 and 25 that you provide is the calculation for row 25...
– yatu
Nov 23 '18 at 11:07
I see what you mean. You might want to specify that the example of rows 15 and 25 that you provide is the calculation for row 25...
– yatu
Nov 23 '18 at 11:07
@Pravat, how do you obtain 0.03 in row 24?
– yatu
Nov 23 '18 at 11:49
@Pravat, how do you obtain 0.03 in row 24?
– yatu
Nov 23 '18 at 11:49
|
show 5 more comments
1 Answer
1
active
oldest
votes
1
Here's a way to do it:
def last_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
first = slice_2min.iloc[0].last_price
return (x.last_price - first)*100/first
def before_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
max_val = slice_2min.last_price.max()
min_val = slice_2min.last_price.min()
return (max_val - min_val)*100/min_val
df = df.set_index(df.time).drop(['time'], axis = 1)
indices = df.loc[df.index.max() - datetime.timedelta(minutes=1):].index.unique()
df_ = df.reset_index()
df_.loc[df_.time.isin(indices), 'last_2mins '] =
df.loc[indices].apply(lambda x: last_2mins(x, df), axis = 1).values
df_.loc[df_.time.isin(indices), 'before_2mins'] =
df.loc[indices].apply(lambda x: before_2mins(x, df), axis = 1).values
time last_price last_2mins before_2mins
0 2018-11-23 10:09:00 110.5 NaN NaN
1 2018-11-23 10:09:00 111.6 NaN NaN
2 2018-11-23 10:09:00 111.5 NaN NaN
3 2018-11-23 10:09:00 112.0 NaN NaN
4 2018-11-23 10:10:00 112.0 NaN NaN
5 2018-11-23 10:10:00 112.6 NaN NaN
6 2018-11-23 10:10:00 112.5 NaN NaN
7 2018-11-23 10:10:00 113.1 NaN NaN
8 2018-11-23 10:11:00 114.3 NaN NaN
9 2018-11-23 10:11:00 114.5 NaN NaN
10 2018-11-23 10:11:00 115.7 NaN NaN
11 2018-11-23 10:12:00 116.5 NaN NaN
12 2018-11-23 10:12:00 116.3 NaN NaN
13 2018-11-23 10:12:00 116.2 NaN NaN
14 2018-11-23 10:13:00 116.5 NaN NaN
15 2018-11-23 10:13:00 117.8 NaN NaN
16 2018-11-23 10:13:00 117.9 NaN NaN
17 2018-11-23 10:14:00 117.5 0.858369 2.323580
18 2018-11-23 10:14:00 118.7 1.888412 2.323580
19 2018-11-23 10:14:00 118.9 2.060086 2.323580
20 2018-11-23 10:14:00 118.3 1.545064 2.323580
21 2018-11-23 10:15:00 118.5 1.716738 2.832618
22 2018-11-23 10:15:00 119.6 2.660944 2.832618
23 2018-11-23 10:15:00 119.5 2.575107 2.832618
24 2018-11-23 10:15:00 119.8 2.832618 2.832618
I will dedicate some time later on to add comments. Let me know if you are sure about the your content in Price_Range_before_2mins. On the meantime, hope this helps.
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Okay @Pravat should be okay now. See if you can tweek the functionbefore_2minsif it is not doing what you want, as I didn't quite understand what you were asking for
– yatu
Nov 23 '18 at 14:39
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Here's a way to do it:
def last_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
first = slice_2min.iloc[0].last_price
return (x.last_price - first)*100/first
def before_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
max_val = slice_2min.last_price.max()
min_val = slice_2min.last_price.min()
return (max_val - min_val)*100/min_val
df = df.set_index(df.time).drop(['time'], axis = 1)
indices = df.loc[df.index.max() - datetime.timedelta(minutes=1):].index.unique()
df_ = df.reset_index()
df_.loc[df_.time.isin(indices), 'last_2mins '] =
df.loc[indices].apply(lambda x: last_2mins(x, df), axis = 1).values
df_.loc[df_.time.isin(indices), 'before_2mins'] =
df.loc[indices].apply(lambda x: before_2mins(x, df), axis = 1).values
time last_price last_2mins before_2mins
0 2018-11-23 10:09:00 110.5 NaN NaN
1 2018-11-23 10:09:00 111.6 NaN NaN
2 2018-11-23 10:09:00 111.5 NaN NaN
3 2018-11-23 10:09:00 112.0 NaN NaN
4 2018-11-23 10:10:00 112.0 NaN NaN
5 2018-11-23 10:10:00 112.6 NaN NaN
6 2018-11-23 10:10:00 112.5 NaN NaN
7 2018-11-23 10:10:00 113.1 NaN NaN
8 2018-11-23 10:11:00 114.3 NaN NaN
9 2018-11-23 10:11:00 114.5 NaN NaN
10 2018-11-23 10:11:00 115.7 NaN NaN
11 2018-11-23 10:12:00 116.5 NaN NaN
12 2018-11-23 10:12:00 116.3 NaN NaN
13 2018-11-23 10:12:00 116.2 NaN NaN
14 2018-11-23 10:13:00 116.5 NaN NaN
15 2018-11-23 10:13:00 117.8 NaN NaN
16 2018-11-23 10:13:00 117.9 NaN NaN
17 2018-11-23 10:14:00 117.5 0.858369 2.323580
18 2018-11-23 10:14:00 118.7 1.888412 2.323580
19 2018-11-23 10:14:00 118.9 2.060086 2.323580
20 2018-11-23 10:14:00 118.3 1.545064 2.323580
21 2018-11-23 10:15:00 118.5 1.716738 2.832618
22 2018-11-23 10:15:00 119.6 2.660944 2.832618
23 2018-11-23 10:15:00 119.5 2.575107 2.832618
24 2018-11-23 10:15:00 119.8 2.832618 2.832618
I will dedicate some time later on to add comments. Let me know if you are sure about the your content in Price_Range_before_2mins. On the meantime, hope this helps.
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Okay @Pravat should be okay now. See if you can tweek the functionbefore_2minsif it is not doing what you want, as I didn't quite understand what you were asking for
– yatu
Nov 23 '18 at 14:39
|
show 1 more comment
1
Here's a way to do it:
def last_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
first = slice_2min.iloc[0].last_price
return (x.last_price - first)*100/first
def before_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
max_val = slice_2min.last_price.max()
min_val = slice_2min.last_price.min()
return (max_val - min_val)*100/min_val
df = df.set_index(df.time).drop(['time'], axis = 1)
indices = df.loc[df.index.max() - datetime.timedelta(minutes=1):].index.unique()
df_ = df.reset_index()
df_.loc[df_.time.isin(indices), 'last_2mins '] =
df.loc[indices].apply(lambda x: last_2mins(x, df), axis = 1).values
df_.loc[df_.time.isin(indices), 'before_2mins'] =
df.loc[indices].apply(lambda x: before_2mins(x, df), axis = 1).values
time last_price last_2mins before_2mins
0 2018-11-23 10:09:00 110.5 NaN NaN
1 2018-11-23 10:09:00 111.6 NaN NaN
2 2018-11-23 10:09:00 111.5 NaN NaN
3 2018-11-23 10:09:00 112.0 NaN NaN
4 2018-11-23 10:10:00 112.0 NaN NaN
5 2018-11-23 10:10:00 112.6 NaN NaN
6 2018-11-23 10:10:00 112.5 NaN NaN
7 2018-11-23 10:10:00 113.1 NaN NaN
8 2018-11-23 10:11:00 114.3 NaN NaN
9 2018-11-23 10:11:00 114.5 NaN NaN
10 2018-11-23 10:11:00 115.7 NaN NaN
11 2018-11-23 10:12:00 116.5 NaN NaN
12 2018-11-23 10:12:00 116.3 NaN NaN
13 2018-11-23 10:12:00 116.2 NaN NaN
14 2018-11-23 10:13:00 116.5 NaN NaN
15 2018-11-23 10:13:00 117.8 NaN NaN
16 2018-11-23 10:13:00 117.9 NaN NaN
17 2018-11-23 10:14:00 117.5 0.858369 2.323580
18 2018-11-23 10:14:00 118.7 1.888412 2.323580
19 2018-11-23 10:14:00 118.9 2.060086 2.323580
20 2018-11-23 10:14:00 118.3 1.545064 2.323580
21 2018-11-23 10:15:00 118.5 1.716738 2.832618
22 2018-11-23 10:15:00 119.6 2.660944 2.832618
23 2018-11-23 10:15:00 119.5 2.575107 2.832618
24 2018-11-23 10:15:00 119.8 2.832618 2.832618
I will dedicate some time later on to add comments. Let me know if you are sure about the your content in Price_Range_before_2mins. On the meantime, hope this helps.
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Okay @Pravat should be okay now. See if you can tweek the functionbefore_2minsif it is not doing what you want, as I didn't quite understand what you were asking for
– yatu
Nov 23 '18 at 14:39
|
show 1 more comment
1
1
1
Here's a way to do it:
def last_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
first = slice_2min.iloc[0].last_price
return (x.last_price - first)*100/first
def before_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
max_val = slice_2min.last_price.max()
min_val = slice_2min.last_price.min()
return (max_val - min_val)*100/min_val
df = df.set_index(df.time).drop(['time'], axis = 1)
indices = df.loc[df.index.max() - datetime.timedelta(minutes=1):].index.unique()
df_ = df.reset_index()
df_.loc[df_.time.isin(indices), 'last_2mins '] =
df.loc[indices].apply(lambda x: last_2mins(x, df), axis = 1).values
df_.loc[df_.time.isin(indices), 'before_2mins'] =
df.loc[indices].apply(lambda x: before_2mins(x, df), axis = 1).values
time last_price last_2mins before_2mins
0 2018-11-23 10:09:00 110.5 NaN NaN
1 2018-11-23 10:09:00 111.6 NaN NaN
2 2018-11-23 10:09:00 111.5 NaN NaN
3 2018-11-23 10:09:00 112.0 NaN NaN
4 2018-11-23 10:10:00 112.0 NaN NaN
5 2018-11-23 10:10:00 112.6 NaN NaN
6 2018-11-23 10:10:00 112.5 NaN NaN
7 2018-11-23 10:10:00 113.1 NaN NaN
8 2018-11-23 10:11:00 114.3 NaN NaN
9 2018-11-23 10:11:00 114.5 NaN NaN
10 2018-11-23 10:11:00 115.7 NaN NaN
11 2018-11-23 10:12:00 116.5 NaN NaN
12 2018-11-23 10:12:00 116.3 NaN NaN
13 2018-11-23 10:12:00 116.2 NaN NaN
14 2018-11-23 10:13:00 116.5 NaN NaN
15 2018-11-23 10:13:00 117.8 NaN NaN
16 2018-11-23 10:13:00 117.9 NaN NaN
17 2018-11-23 10:14:00 117.5 0.858369 2.323580
18 2018-11-23 10:14:00 118.7 1.888412 2.323580
19 2018-11-23 10:14:00 118.9 2.060086 2.323580
20 2018-11-23 10:14:00 118.3 1.545064 2.323580
21 2018-11-23 10:15:00 118.5 1.716738 2.832618
22 2018-11-23 10:15:00 119.6 2.660944 2.832618
23 2018-11-23 10:15:00 119.5 2.575107 2.832618
24 2018-11-23 10:15:00 119.8 2.832618 2.832618
I will dedicate some time later on to add comments. Let me know if you are sure about the your content in Price_Range_before_2mins. On the meantime, hope this helps.
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
Here's a way to do it:
def last_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
first = slice_2min.iloc[0].last_price
return (x.last_price - first)*100/first
def before_2mins(x, df):
from_time = x.name - datetime.timedelta(minutes=2)
slice_2min = df.loc[from_time:x.name,:]
max_val = slice_2min.last_price.max()
min_val = slice_2min.last_price.min()
return (max_val - min_val)*100/min_val
df = df.set_index(df.time).drop(['time'], axis = 1)
indices = df.loc[df.index.max() - datetime.timedelta(minutes=1):].index.unique()
df_ = df.reset_index()
df_.loc[df_.time.isin(indices), 'last_2mins '] =
df.loc[indices].apply(lambda x: last_2mins(x, df), axis = 1).values
df_.loc[df_.time.isin(indices), 'before_2mins'] =
df.loc[indices].apply(lambda x: before_2mins(x, df), axis = 1).values
time last_price last_2mins before_2mins
0 2018-11-23 10:09:00 110.5 NaN NaN
1 2018-11-23 10:09:00 111.6 NaN NaN
2 2018-11-23 10:09:00 111.5 NaN NaN
3 2018-11-23 10:09:00 112.0 NaN NaN
4 2018-11-23 10:10:00 112.0 NaN NaN
5 2018-11-23 10:10:00 112.6 NaN NaN
6 2018-11-23 10:10:00 112.5 NaN NaN
7 2018-11-23 10:10:00 113.1 NaN NaN
8 2018-11-23 10:11:00 114.3 NaN NaN
9 2018-11-23 10:11:00 114.5 NaN NaN
10 2018-11-23 10:11:00 115.7 NaN NaN
11 2018-11-23 10:12:00 116.5 NaN NaN
12 2018-11-23 10:12:00 116.3 NaN NaN
13 2018-11-23 10:12:00 116.2 NaN NaN
14 2018-11-23 10:13:00 116.5 NaN NaN
15 2018-11-23 10:13:00 117.8 NaN NaN
16 2018-11-23 10:13:00 117.9 NaN NaN
17 2018-11-23 10:14:00 117.5 0.858369 2.323580
18 2018-11-23 10:14:00 118.7 1.888412 2.323580
19 2018-11-23 10:14:00 118.9 2.060086 2.323580
20 2018-11-23 10:14:00 118.3 1.545064 2.323580
21 2018-11-23 10:15:00 118.5 1.716738 2.832618
22 2018-11-23 10:15:00 119.6 2.660944 2.832618
23 2018-11-23 10:15:00 119.5 2.575107 2.832618
24 2018-11-23 10:15:00 119.8 2.832618 2.832618
I will dedicate some time later on to add comments. Let me know if you are sure about the your content in Price_Range_before_2mins. On the meantime, hope this helps.
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
edited Nov 23 '18 at 14:38
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
answered Nov 23 '18 at 12:37
yatuyatu
6,1931725
6,1931725
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Okay @Pravat should be okay now. See if you can tweek the functionbefore_2minsif it is not doing what you want, as I didn't quite understand what you were asking for
– yatu
Nov 23 '18 at 14:39
|
show 1 more comment
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Okay @Pravat should be okay now. See if you can tweek the functionbefore_2minsif it is not doing what you want, as I didn't quite understand what you were asking for
– yatu
Nov 23 '18 at 14:39
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Thanks for the code. Yea I want the percentage range of the price before 2 mins for a period of 2 mins as per the the question i.e. (max value of last_price from row 05 to row 14) - (min value of last_price from row 05 to row 14)*100/ (min value of last_price from row 05 to row 14)
– Pravat
Nov 23 '18 at 12:44
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Just realized there a change of var name that is did not update. Will edit later
– yatu
Nov 23 '18 at 12:50
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Sir, in last_2mins function getting error on "first" - Not Defined
– Pravat
Nov 23 '18 at 13:10
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Yes it's what I was mentioning, will fix as soon as I get to use my laptop
– yatu
Nov 23 '18 at 13:18
Okay @Pravat should be okay now. See if you can tweek the function
before_2minsif it is not doing what you want, as I didn't quite understand what you were asking for– yatu
Nov 23 '18 at 14:39
Okay @Pravat should be okay now. See if you can tweek the function
before_2minsif it is not doing what you want, as I didn't quite understand what you were asking for– yatu
Nov 23 '18 at 14:39
|
show 1 more comment
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2
Have you tried to solve these yourself? Can you share with us some attempts at solving it?
– yatu
Nov 23 '18 at 10:50
I think it will help you stackoverflow.com/questions/12376863/… use a.apply(command)
– Jony Kim
Nov 23 '18 at 10:56
Sir, I am new to python and unable to handle data with calculating Time Difference.
– Pravat
Nov 23 '18 at 11:06
I see what you mean. You might want to specify that the example of rows 15 and 25 that you provide is the calculation for row 25...
– yatu
Nov 23 '18 at 11:07
@Pravat, how do you obtain 0.03 in row 24?
– yatu
Nov 23 '18 at 11:49