I am trying to work out the time delta between values in a grouped pandas df.
My df looks like this:
Location ID Item Qty Time
0 7 202545942 100130 1 07:19:46
1 8 202545943 100130 1 07:20:08
2 11 202545950 100130 1 07:20:31
3 13 202545955 100130 1 07:21:08
4 15 202545958 100130 1 07:21:18
5 18 202545963 100130 3 07:21:53
6 217 202546320 100130 1 07:22:43
7 219 202546324 100130 1 07:22:54
8 229 202546351 100130 1 07:23:32
9 246 202546376 100130 1 07:24:09
10 273 202546438 100130 1 07:24:37
11 286 202546464 100130 1 07:24:59
12 296 202546490 100130 1 07:25:16
13 297 202546491 100130 1 07:25:24
14 310 202546516 100130 1 07:25:59
15 321 202546538 100130 1 07:26:17
16 329 202546549 100130 1 07:28:09
17 388 202546669 100130 1 07:29:02
18 420 202546717 100130 2 07:30:01
19 451 202546766 100130 1 07:30:19
20 456 202546773 100130 1 07:30:27
(...)
42688 458 202546777 999969 1 06:51:16
42689 509 202546884 999969 1 06:53:09
42690 567 202546977 999969 1 06:54:21
42691 656 202547104 999969 1 06:57:27
I have grouped this using the following method:
ndf = df.groupby(['ID','Location','Time'])
If I add .size() to the end of the above and print(ndf) I get the following output:
(...)
ID Location Time
995812 696 07:10:36 1
730 07:11:41 1
761 07:12:30 1
771 07:20:49 1
995820 381 06:55:07 1
761 07:12:44 1
(...)
This is the as desired.
My challenge is that I need to work out the time delta between each time per Item and add this as a column in the dataframe grouping. It should give me the following:
ID Location Time Delta
(...)
995812 696 07:10:36 0
730 07:11:41 00:01:05
761 07:12:30 00:00:49
771 07:20:49 00:08:19
995820 381 06:55:07 0
761 07:12:44 00:17:37
(...)
I am pulling my hair out trying to work out a method of doing this, so I'm turning to the greats.
Please help. Thanks in advance.
Convert Time column to timedeltas by to_timedelta, sort by all 3 columns by DataFrame.sort_values, get difference per groups by DataFrameGroupBy.diff, replace missing values to 0 timedelta by Series.fillna:
#if strings astype should be omit
df['Time'] = pd.to_timedelta(df['Time'].astype(str))
df = df.sort_values(['ID','Location','Time'])
df['Delta'] = df.groupby('ID')['Time'].diff().fillna(pd.Timedelta(0))
Also is possible convert timedeltas to seconds - add Series.dt.total_seconds:
df['Delta_sec'] = df.groupby('ID')['Time'].diff().dt.total_seconds().fillna(0)
If you just wanted to iterate over the groupby object, based on your original question title you can do it:
for (x, y) in df.groupby(['ID','Location','Time']):
print("{0}, {1}".format(x, y))
# your logic
However, this works for 10.000 rows, 100.000 rows, but not so good for 10^6 rows or more.
Related
I imported my data file and isolated the first letter of each word, and provided the count of the word. My next step is to sort the letters in ascending order 'a-z'. This is the code that I have right now:
import pandas as pd
df = pd.read_csv(text.txt", names=['FirstNames'])
df
df['FirstLetter'] = df['FirstNames'].str[:1]
df
df['FirstLetter'] = df['FirstLetter'].str.lower()
df
df['FirstLetter'].value_counts()
df
df2 = df['FirstLetter'].index.value_counts()
df2
Using .index.value_counts() wasn't working for me. It turned this output:
Out[72]:
2047 1
4647 1
541 1
4639 1
2592 1
545 1
4643 1
2596 1
549 1
2600 1
2612 1
553 1
4651 1
2604 1
557 1
4655 1
2608 1
561 1
2588 1
4635 1
..
`````````
How can I fix this?
You can use the sort_index() function. This should work df['FirstLetter'].value_counts().sort_index()
Suppose, you have a column in excel, with values like this... there are only 5500 numbers present but it show length 5602 means that 102 strings are present
4 SELECTIO
6 N NO
14 37001
26 37002
38 37003
47 37004
60 37005
73 37006
82 37007
92 37008
105 37009
119 37010
132 37011
143 37012
157 37013
168 37014
184 37015
196 37016
207 37017
220 37018
236 37019
253 37020
267 37021
280 37022
287 Krishan
290 37023
300 37024
316 37025
337 37026
365 37027
...
74141 42471
74154 42472
74169 42473
74184 42474
74200 42475
74216 42476
74233 42477
74242 42478
74256 42479
74271 42480
74290 42481
74309 42482
74323 42483
74336 42484
74350 42485
74365 42486
74378 42487
74389 42488
74398 42489
74413 42490
74430 42491
74446 42492
74459 42493
74474 42494
74491 42495
74504 42496
74516 42497
74530 42498
74544 42499
74558 42500
Name: Selection No., Length: 5602, dtype: object
and I want to get only numeric values like this in python using pandas
37001
37002
37003
37004
37005
how can I do this? I have attached my code in python using pandas..............................................
def selection(sle):
if sle in re.match('[3-4][0-9]{4}',sle):
return 1
else:
return 0
select['status'] = select['Selection No.'].apply(selection)
and now I am geting an "argument of type 'NoneType' is not iterable" error.
Try using Numpy with np.isreal and only select numbers..
import pandas as pd
import numpy as np
df = pd.DataFrame({'SELECTIO':['N NO',37002,37003,'Krishan',37004,'singh',37005], 'some_col':[4,6,14,26,38,47,60]})
df
SELECTIO some_col
0 N NO 4
1 37002 6
2 37003 14
3 Krishan 26
4 37004 38
5 singh 47
6 37005 60
>>> df[df[['SELECTIO']].applymap(np.isreal).all(1)]
SELECTIO some_col
1 37002 6
2 37003 14
4 37004 38
6 37005 60
result:
Specific to column SELECTIO ..
df[df[['SELECTIO']].applymap(np.isreal).all(1)]
SELECTIO some_col
1 37002 6
2 37003 14
4 37004 38
6 37005 60
OR just another approach importing numbers + lambda :
import numbers
df[df[['SELECTIO']].applymap(lambda x: isinstance(x, numbers.Number)).all(1)]
SELECTIO some_col
1 37002 6
2 37003 14
4 37004 38
6 37005 60
Note: there is problem when you are extracting a column you are using ['Selection No.'] but indeed you have a Space in the name it will be like ['Selection No. '] that's the reason you are getting KeyError while executing it, try and see!
Your function contains wrong expression: if sle in re.match('[3-4][0-9]{4}',sle): - it tries to find a column value sle IN match object which "always have a boolean value of True" (re.match returns None when there's no match)
I would suggest to proceed with pd.Series.str.isnumeric function:
In [544]: df
Out[544]:
Selection No.
0 37001
1 37002
2 37003
3 asnsh
4 37004
5 singh
6 37005
In [545]: df['Status'] = df['Selection No.'].str.isnumeric().astype(int)
In [546]: df
Out[546]:
Selection No. Status
0 37001 1
1 37002 1
2 37003 1
3 asnsh 0
4 37004 1
5 singh 0
6 37005 1
If a strict regex pattern is required - use pd.Series.str.contains function:
df['Status'] = df['Selection No.'].str.contains('^[3-4][0-9]{4}$', regex=True).astype(int)
I grouped my data by month. Now I need to know at which observation/index my group starts and ends.
What I have is the following output where the second column represents the number of observation in each month:
date
01 145
02 2232
03 12785
04 16720
Name: date, dtype: int64
with this code:
leave.groupby([leave['date'].dt.strftime('%m')])['date'].count()
What I want though is an index range I could access later. Somehow like that (the format doesn't really matter and I don't mind if it returns a list or a data frame)
date
01 0 - 145
02 146 - 2378
03 2378 - 15163
04 15164 - 31884
try the following - using shift
df['data'] = df['data'].shift(1).add(1).fillna(0).apply(int).apply(str) + ' - ' + df['data'].apply(str)
OUTPUT:
data
date
1 0 - 145
2 146 - 2232
3 2233 - 12785
4 12786 - 16720
5 16721 - 30386
6 30387 - 120157
I think you are asking for a data frame containing the indices of first and last occurrences of each value.
How about something like this.
Example data (note -- it's better to include reproducible data in your question so I don't have to guess):
import pandas as pd
import numpy as np
np.random.seed(123)
n = 500
df = pd.DataFrame(
{'date':pd.to_datetime(
pd.DataFrame( { 'year': np.random.choice(range(2017,2019), size=n),
'month': np.random.choice(range(1,13), size=n),
'day': np.random.choice(range(1,28), size=n)
} )
) }
)
Approach:
pd.DataFrame( ( { '_month_':x,'firstIndex':y[0],'lastIndex':y[-1]}
for x, y in df.index.groupby(df['date'].dt.month).items()
)
)
Result:
_month_ firstIndex lastIndex
0 1 0 495
1 2 21 499
2 3 1 488
3 4 5 498
4 5 14 492
5 6 12 470
6 7 15 489
7 8 2 494
8 9 18 475
9 10 3 491
10 11 10 473
11 12 7 497
If you are only going use it for indexing in a loop, you wouldn't have to wrap it in pd.DataFrame() -- you could just leave it as a generator.
The Scenario
My dataset was in format as follows:
Which I refer as ACTUAL FORMAT
uid iid rat tmp
196 242 3 881250949
186 302 3 891717742
22 377 1 878887116
244 51 2 880606923
166 346 1 886397596
298 474 4 884182806
115 265 2 881171488
253 465 5 891628467
305 451 3 886324817
6 86 3 883603013
and while passing it to other function (KMeans Clustering) it requires to be format like this, which I've created using Pivot mapping:
Which I refer as MATRIX FORMAT
uid 1 2 3 4
4 4.3320762062 4.3407749532 4.3111995162 4.3411425423
5 4 3 2.1952622349 3.1913491995
6 4 3.4233243638 3.8255108621 3.948791424
7 4.4983411706 4.0477240538 4.0241460801 5
8 4.1773004578 4.0191412859 4.0442369862 4.1754642909
9 4.2733984521 4.2797130861 4.2682723131 4.2816986988
15 1 3.0554789259 3.2279546684 3.1282278957
16 5 4.3473697565 4.0675394438 5
The Problem:
Now, Since I need the result / MATRIX FORMAT Data to passed again to the First Algorithm, I need to convert it to OLD FORMAT.
Coversion:
For conversion of OLD to MATRIX Format I did:
Pivot_Matrix = source_data.pivot(values='rat', index='uid', columns='iid')
I tried reversing & interchanging of values to get the OLD FORMAT, which has apparently failed. Is there any way to retrieve MATRIX to OLD FORMAT?
You need stack with rename_axis for columns names and last reset_index:
df = df.stack().rename_axis(('uid','iid')).reset_index(name='rat')
print (df.head())
uid iid rat
0 4 1 4.332076
1 4 2 4.340775
2 4 3 4.311200
3 4 4 4.341143
4 5 1 4.000000
I met a problem in formatting pivot table that created by Pandas.
So I made a matrix table between 2 columns (A,B) from my source data, by using pandas.pivot_table with A as Column, and B as Index.
>> df = PD.read_excel("data.xls")
>> table = PD.pivot_table(df,index=["B"],
values='Count',columns=["A"],aggfunc=[NUM.sum],
fill_value=0,margins=True,dropna= True)
>> table
It returns as:
sum
A 1 2 3 All
B
1 23 52 0 75
2 16 35 12 65
3 56 0 0 56
All 95 87 12 196
And I hope to have a format like this:
A All_B
1 2 3
1 23 52 0 75
B 2 16 35 12 65
3 56 0 0 56
All_A 95 87 12 196
How should I do this? Thanks very much ahead.
The table returned by pd.pivot_table is very convenient to do work on (it's single-level index/column) and normally does NOT require any further format manipulation. But if you insist on changing the format to the one you mentioned in the post, then you need to construct a multi-level index/column using pd.MultiIndex. Here is an example on how to do it.
Before manipulation,
import pandas as pd
import numpy as np
np.random.seed(0)
a = np.random.randint(1, 4, 100)
b = np.random.randint(1, 4, 100)
df = pd.DataFrame(dict(A=a,B=b,Val=np.random.randint(1,100,100)))
table = pd.pivot_table(df, index='A', columns='B', values='Val', aggfunc=sum, fill_value=0, margins=True)
print(table)
B 1 2 3 All
A
1 454 649 770 1873
2 628 576 467 1671
3 376 247 481 1104
All 1458 1472 1718 4648
After:
multi_level_column = pd.MultiIndex.from_arrays([['A', 'A', 'A', 'All_B'], [1,2,3,'']])
multi_level_index = pd.MultiIndex.from_arrays([['B', 'B', 'B', 'All_A'], [1,2,3,'']])
table.index = multi_level_index
table.columns = multi_level_column
print(table)
A All_B
1 2 3
B 1 454 649 770 1873
2 628 576 467 1671
3 376 247 481 1104
All_A 1458 1472 1718 4648