I want to produce a heatmap using matplolib and this pandas dataframe:
class time day
0 URxvt 2 6
1 Emacs 3 6
2 Firefox 90 6
3 KeePassXC 5 6
4 URxvt 91 6
.. ... ... ...
144 Matplotlib 1 1
145 Matplotlib 1 1
146 Matplotlib 2 1
147 Matplotlib 5 1
148 Matplotlib 93 1
[149 rows x 3 columns]
I want to produce a heatmap with day (from 0 to 6 (but for the moment 0, 1 and 6)) on x-axis and class on y-axis, values are aggregate sums according to class and day).
I tried to groupby these two columns which produces:
time
class day
Emacs 0 1149
1 130
6 634
Eog 1 83
6 66
Evince 0 775
6 60
File-roller 0 40
Firefox 0 32109
1 6344
6 9887
GParted 1 25
Gedit 0 77
1 7
Gimp-2.10 6 25
Gmpc 1 73
Gnome-disks 1 21
Gtk-recordmydesktop 0 57
Gufw.py 6 100
KeePassXC 0 44
1 17
6 126
Matplotlib 1 151
Org.gnome.Nautilus 0 141
1 559
6 68
Scangearmp2 6 28
Totem 0 12
URxvt 0 346
1 488
6 3364
vlc 0 22
but I can't get a proper heatmap (with X:day Y:class and values: time)
You can try:
sns.heatmap(df.groupby(['class','day'])['time'].sum()
.unstack('day',fill_value=0)
)
Output:
Related
I'm trying to create or generate some graphs in stacked bar I'm using this data:
index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
0 No 94 123 96 108 122 106.0 95.0 124 104 118 73 82 106 124 109 70 59
1 Yes 34 4 33 21 5 25.0 34.0 5 21 9 55 46 21 3 19 59 41
2 Dont know 1 2 1 1 2 NaN NaN 1 4 2 2 2 2 2 2 1 7
Basically I want to use the columns names as x and the Yes, No, Don't know as the Y values, here is my code and the result that I have at the moment.
ax = dfu.plot.bar(x='index', stacked=True)
UPDATE:
Here is an example:
data = [{0:1,1:2,2:3},{0:3,1:2,2:1},{0:1,1:1,2:1}]
index = ["yes","no","dont know"]
df = pd.DataFrame(data,index=index)
df.T.plot.bar(stacked=True) # Note .T is used to transpose the DataFrame
I am trying to append every column from one row into another row, I want to do this for every row, but some row will not have any values, take a look at my code it will be more clear:
Here is my data
date day_of_week day_of_month day_of_year month_of_year
5/1/2017 0 1 121 5
5/2/2017 1 2 122 5
5/3/2017 2 3 123 5
5/4/2017 3 4 124 5
5/8/2017 0 8 128 5
5/9/2017 1 9 129 5
5/10/2017 2 10 130 5
5/11/2017 3 11 131 5
5/12/2017 4 12 132 5
5/15/2017 0 15 135 5
5/16/2017 1 16 136 5
5/17/2017 2 17 137 5
5/18/2017 3 18 138 5
5/19/2017 4 19 139 5
5/23/2017 1 23 143 5
5/24/2017 2 24 144 5
5/25/2017 3 25 145 5
5/26/2017 4 26 146 5
Here is my current code:
s = df_md['date'].shift(-1)
df_md['next_calendarday'] = s.mask(s.dt.dayofweek.diff().lt(0))
df_md.set_index('date', inplace=True)
df_md.apply(lambda row: GetNextDayMarketData(row, df_md), axis=1)
def GetNextDayMarketData(row, dataframe):
if(row['next_calendarday'] is pd.NaT):
return
key = row['next_calendarday'].strftime("%Y-%m-%d")
nextrow = dataframe.loc[key]
for index, val in nextrow.iteritems():
if(index != "next_calendarday"):
dataframe.loc[row.name, index+'_nextday'] = val
This works but it's so slow it might as well not work. Here is what the result should look like, you can see that the value from the next row has been added to the previous row. The kicker is that it's the next calendar date and not just the next row in the sequence. If a row does not have an entry for next calendar date, it will simply be blank.
Here is the expected result in csv
date day_of_week day_of_month day_of_year month_of_year next_workingday day_of_week_nextday day_of_month_nextday day_of_year_nextday month_of_year_nextday
5/1/2017 0 1 121 5 5/2/2017 1 2 122 5
5/2/2017 1 2 122 5 5/3/2017 2 3 123 5
5/3/2017 2 3 123 5 5/4/2017 3 4 124 5
5/4/2017 3 4 124 5
5/8/2017 0 8 128 5 5/9/2017 1 9 129 5
5/9/2017 1 9 129 5 5/10/2017 2 10 130 5
5/10/2017 2 10 130 5 5/11/2017 3 11 131 5
5/11/2017 3 11 131 5 5/12/2017 4 12 132 5
5/12/2017 4 12 132 5
5/15/2017 0 15 135 5 5/16/2017 1 16 136 5
5/16/2017 1 16 136 5 5/17/2017 2 17 137 5
5/17/2017 2 17 137 5 5/18/2017 3 18 138 5
5/18/2017 3 18 138 5 5/19/2017 4 19 139 5
5/19/2017 4 19 139 5
5/23/2017 1 23 143 5 5/24/2017 2 24 144 5
5/24/2017 2 24 144 5 5/25/2017 3 25 145 5
5/25/2017 3 25 145 5 5/26/2017 4 26 146 5
5/26/2017 4 26 146 5
5/30/2017 1 30 150 5
Use DataFrame.join with remove column next_calendarday_nextday:
df = df.set_index('date')
df = (df.join(df, on='next_calendarday', rsuffix='_nextday')
.drop('next_calendarday_nextday', axis=1))
I have an set of stock information, with datetime set as index, stock market only open on weekdays so all my rows are weekdays, which is fine, I would like to determine if a row is start of the week or end of week, which might NOT always fall on Monday/Friday due to holidays. A better idea is to determine if there is an row entry on the next/previous day in the dataframe ( since my data is guaranteed to only exist for workday), but I dont know how to calculate this. Here is an example of my data:
date day_of_week day_of_month day_of_year month_of_year
5/1/2017 0 1 121 5
5/2/2017 1 2 122 5
5/3/2017 2 3 123 5
5/4/2017 3 4 124 5
5/8/2017 0 8 128 5
5/9/2017 1 9 129 5
5/10/2017 2 10 130 5
5/11/2017 3 11 131 5
5/12/2017 4 12 132 5
5/15/2017 0 15 135 5
5/16/2017 1 16 136 5
5/17/2017 2 17 137 5
5/18/2017 3 18 138 5
5/19/2017 4 19 139 5
5/23/2017 1 23 143 5
5/24/2017 2 24 144 5
5/25/2017 3 25 145 5
5/26/2017 4 26 146 5
5/30/2017 1 30 150 5
Here is my current code
# Date fields
def DateFields(df_input):
dates = df_input.index.to_series()
df_input['day_of_week'] = dates.dt.dayofweek
df_input['day_of_month'] = dates.dt.day
df_input['day_of_year'] = dates.dt.dayofyear
df_input['month_of_year'] = dates.dt.month
df_input['isWeekStart'] = "No" #<--- Need help here
df_input['isWeekEnd'] = "No" #<--- Need help here
df_input['date'] = dates.dt.strftime('%Y-%m-%d')
return df_input
How can I calculate if a row is beginning of week and end of week?
Example of what I am looking for:
date day_of_week day_of_month day_of_year month_of_year isWeekStart isWeekEnd
5/1/2017 0 1 121 5 1 0
5/2/2017 1 2 122 5 0 0
5/3/2017 2 3 123 5 0 0
5/4/2017 3 4 124 5 0 1 # short week, Thursday is last work day
5/8/2017 0 8 128 5 1 0
5/9/2017 1 9 129 5 0 0
5/10/2017 2 10 130 5 0 0
5/11/2017 3 11 131 5 0 0
5/12/2017 4 12 132 5 0 1
5/15/2017 0 15 135 5 1 0
5/16/2017 1 16 136 5 0 0
5/17/2017 2 17 137 5 0 0
5/18/2017 3 18 138 5 0 0
5/19/2017 4 19 139 5 0 1
5/23/2017 1 23 143 5 1 0 # short week, Tuesday is first work day
5/24/2017 2 24 144 5 0 0
5/25/2017 3 25 145 5 0 0
5/26/2017 4 26 146 5 0 1
5/30/2017 1 30 150 5 1 0
EDIT: I forgot that some holidays fall during the middle of week, in this situation, it would be good if it can treat these as a separate "week" with before and after marked accordingly. Although if it's not smart enough to figure this out, just getting the long weekend would be a good start.
Here's an idea with BusinessDay:
prev_working_day = df['date'] - pd.tseries.offsets.BusinessDay(1)
df['isFirstWeekDay'] = (df['date'].dt.isocalendar().week !=
prev_working_day.dt.isocalendar().week)
And similar for last business day. Note that the default holiday calendar is US'. Check out this post for a different one.
Output:
date day_of_week day_of_month day_of_year month_of_year isFirstWeekDay
0 2017-05-01 0 1 121 5 True
1 2017-05-02 1 2 122 5 False
2 2017-05-03 2 3 123 5 False
3 2017-05-04 3 4 124 5 False
4 2017-05-08 0 8 128 5 True
5 2017-05-09 1 9 129 5 False
6 2017-05-10 2 10 130 5 False
7 2017-05-11 3 11 131 5 False
8 2017-05-12 4 12 132 5 False
9 2017-05-15 0 15 135 5 True
10 2017-05-16 1 16 136 5 False
11 2017-05-17 2 17 137 5 False
12 2017-05-18 3 18 138 5 False
13 2017-05-19 4 19 139 5 False
14 2017-05-23 1 23 143 5 False
15 2017-05-24 2 24 144 5 False
16 2017-05-25 3 25 145 5 False
17 2017-05-26 4 26 146 5 False
18 2017-05-30 1 30 150 5 False
Here's an approach using weekly groupby.
df['date'] = pd.to_datetime(df['date'])
business_days = df.assign(date_copy = df['date']).groupby(pd.Grouper(key='date_copy', freq='W'))['date'].apply(list).to_frame()
business_days['isWeekStart'] = business_days['date'].apply(lambda x: [1 if i == min(x) else 0 for i in x])
business_days['isWeekEnd'] = business_days['date'].apply(lambda x: [1 if i == max(x) else 0 for i in x])
business_days = business_days.apply(pd.Series.explode)
pd.merge(df, business_days, left_on='date', right_on='date')
output:
date day_of_week day_of_month day_of_year month_of_year isWeekStart isWeekEnd
0 2017-05-01 0 1 121 5 1 0
1 2017-05-02 1 2 122 5 0 0
2 2017-05-03 2 3 123 5 0 0
3 2017-05-04 3 4 124 5 0 1
4 2017-05-08 0 8 128 5 1 0
5 2017-05-09 1 9 129 5 0 0
6 2017-05-10 2 10 130 5 0 0
7 2017-05-11 3 11 131 5 0 0
8 2017-05-12 4 12 132 5 0 1
9 2017-05-15 0 15 135 5 1 0
10 2017-05-16 1 16 136 5 0 0
11 2017-05-17 2 17 137 5 0 0
12 2017-05-18 3 18 138 5 0 0
13 2017-05-19 4 19 139 5 0 1
14 2017-05-23 1 23 143 5 1 0
15 2017-05-24 2 24 144 5 0 0
16 2017-05-25 3 25 145 5 0 0
17 2017-05-26 4 26 146 5 0 1
18 2017-05-30 1 30 150 5 1 1
Note that 2017-05-30 is marked as both WeekStart and WeekEnd because it is the only date of that week.
Say I have a DataFrame full of positive samples and context features for a given user:
target user cashtag sector industry
0 1 170 4979 3 70
1 1 170 5539 3 70
2 1 170 7271 3 70
3 1 170 7428 3 70
4 1 170 686 7 139
where a positive sample is a user having interacted with a cashtag and is denoted by target = 1.
What is a quick way for me to generate negative samples in the ratio 1:2 (+ve:-ve) for each interaction, denoted by target = -1?
EDIT: Sample for clarity below (for the first two positive samples)
target user cashtag sector industry
0 1 170 4979 3 70
1 -1 170 3224 7 181
2 -1 170 4331 7 180
3 1 170 5539 3 70
4 -1 170 9304 4 59
5 -1 170 3833 6 185
For instance, for each cashtag a user has interacted with, I'd like to pick at random 2 other cashtags that they haven't interacted with and add them as negative samples to the dataframe; effectively increasing the size of the dataframe to 3 times its original size.
It would also be helpful to check if the negative sample hasn't already been entered for that user, cashtag combination.
Here my solution:
data="""
target user cashtag sector industry
1 170 4979 3 70
1 170 5539 3 70
1 170 7271 3 70
1 170 7428 3 70
1 170 686 7 139
"""
df = pd.read_csv(pd.compat.StringIO(data), sep='\s+')
df1 = pd.DataFrame(columns = df.columns)
cashtag = df['cashtag'].values.tolist()
#function to randomize some numbers
def randomnumber(v):
return np.random.randint(v, size=1)
def addNewRow(x):
for i in range(2): #add 2 new rows
cash = cashtag[0]
while cash in cashtag: #check if cashtag already used
cash = randomnumber(5000)[0] #random number between 0 and 5000
cashtag.append(cash)
sector = randomnumber(10)[0]
industry = randomnumber(200)[0]
df1.loc[df1.shape[0]] = [-1, x.user, cash, sector, industry]
df.apply(lambda x: addNewRow(x), axis=1)
df = df.append(df1).reset_index()
print(df)
output:
index target user cashtag sector industry
0 0 1 170 4979 3 70
1 1 1 170 5539 3 70
2 2 1 170 7271 3 70
3 3 1 170 7428 3 70
4 4 1 170 686 7 139
5 0 -1 170 544 2 59
6 1 -1 170 3202 8 165
7 2 -1 170 2673 0 40
8 3 -1 170 4021 1 30
9 4 -1 170 682 6 3
10 5 -1 170 2446 1 80
11 6 -1 170 4026 9 193
12 7 -1 170 4070 9 197
13 8 -1 170 2900 1 57
14 9 -1 170 3287 0 21
The new random rows are put at the end of dataframe
Given the following dataframe df:
app platform uuid minutes
0 1 0 a696ccf9-22cb-428b-adee-95c9a97a4581 67
1 2 0 8e17a2eb-f0ee-49ae-b8c2-c9f9926aa56d 1
2 2 1 40AD6CD1-4A7B-48DD-8815-1829C093A95C 13
3 1 0 26c1022a-7a8e-42a2-b7cc-bea6bffa7a6f 2
4 2 0 34271596-eebb-4423-b890-dc3761ed37ca 8
5 3 1 C57D0F52-B565-4322-85D2-C2798F7CA6FF 16
6 2 0 245501ec2e39cb782bab1fb02d7813b7 1
7 3 1 DE6E4714-5A3C-4C80-BD81-EAACB2364DF0 30
8 3 0 f88eb774-fdf3-4d1d-a91d-0b4ab95cf36e 10
9 2 0 9c08c860-7a6d-4810-a5c3-f3af2a3fcf66 470
10 3 1 19fdaedfd0dbdaf6a7a6b49619f11a19 3
11 3 1 AAF1CFF7-4564-4C79-B2D8-F0AAF9C9971B 58
12 2 0 4eb1024b-c293-42a4-95a2-31b20c3b524b 24
13 3 1 8E0B0BE3-8553-4F38-9837-6C907E01F84C 7
14 3 1 E8B2849C-F050-4DCD-B311-5D57015466AE 465
15 2 0 ec7fedb6-b118-424a-babe-b8ffad579685 266
16 1 0 7e302dcb-ceaf-406c-a9e5-66933d921064 184
17 2 0 f786528ded200c9f553dd3a5e9e9bb2d 10
18 3 1 1E291633-AF27-4DFB-8DA4-4A5B63F175CF 13
19 2 0 953a525c-97e0-4c2f-90e0-dfebde3ec20d 2408`
I'll group it:
y=df.groupby(['app','platform','uuid']).sum().reset_index().sort(['app','platform','minutes'],ascending=[1,1,0]).set_index(['app','platform','uuid'])
minutes
app platform uuid
1 0 7e302dcb-ceaf-406c-a9e5-66933d921064 184
a696ccf9-22cb-428b-adee-95c9a97a4581 67
26c1022a-7a8e-42a2-b7cc-bea6bffa7a6f 2
2 0 953a525c-97e0-4c2f-90e0-dfebde3ec20d 2408
9c08c860-7a6d-4810-a5c3-f3af2a3fcf66 470
ec7fedb6-b118-424a-babe-b8ffad579685 266
4eb1024b-c293-42a4-95a2-31b20c3b524b 24
f786528ded200c9f553dd3a5e9e9bb2d 10
34271596-eebb-4423-b890-dc3761ed37ca 8
245501ec2e39cb782bab1fb02d7813b7 1
8e17a2eb-f0ee-49ae-b8c2-c9f9926aa56d 1
1 40AD6CD1-4A7B-48DD-8815-1829C093A95C 13
3 0 f88eb774-fdf3-4d1d-a91d-0b4ab95cf36e 10
1 E8B2849C-F050-4DCD-B311-5D57015466AE 465
AAF1CFF7-4564-4C79-B2D8-F0AAF9C9971B 58
DE6E4714-5A3C-4C80-BD81-EAACB2364DF0 30
C57D0F52-B565-4322-85D2-C2798F7CA6FF 16
1E291633-AF27-4DFB-8DA4-4A5B63F175CF 13
8E0B0BE3-8553-4F38-9837-6C907E01F84C 7
19fdaedfd0dbdaf6a7a6b49619f11a19 3
So that I got its minutes per uuid in decrescent order.
Now, I will sum the cumulative minutes per app/platform/uuid:
y.groupby(level=[0,1]).cumsum()
app platform uuid
1 0 7e302dcb-ceaf-406c-a9e5-66933d921064 184
a696ccf9-22cb-428b-adee-95c9a97a4581 251
26c1022a-7a8e-42a2-b7cc-bea6bffa7a6f 253
2 0 953a525c-97e0-4c2f-90e0-dfebde3ec20d 2408
9c08c860-7a6d-4810-a5c3-f3af2a3fcf66 2878
ec7fedb6-b118-424a-babe-b8ffad579685 3144
4eb1024b-c293-42a4-95a2-31b20c3b524b 3168
f786528ded200c9f553dd3a5e9e9bb2d 3178
34271596-eebb-4423-b890-dc3761ed37ca 3186
245501ec2e39cb782bab1fb02d7813b7 3187
8e17a2eb-f0ee-49ae-b8c2-c9f9926aa56d 3188
1 40AD6CD1-4A7B-48DD-8815-1829C093A95C 13
3 0 f88eb774-fdf3-4d1d-a91d-0b4ab95cf36e 10
1 E8B2849C-F050-4DCD-B311-5D57015466AE 465
AAF1CFF7-4564-4C79-B2D8-F0AAF9C9971B 523
DE6E4714-5A3C-4C80-BD81-EAACB2364DF0 553
C57D0F52-B565-4322-85D2-C2798F7CA6FF 569
1E291633-AF27-4DFB-8DA4-4A5B63F175CF 582
8E0B0BE3-8553-4F38-9837-6C907E01F84C 589
19fdaedfd0dbdaf6a7a6b49619f11a19 592
My question is: how can I get the percent agains the total cumulative sum, per group, i.e, something like this:
app platform uuid
1 0 7e302dcb-ceaf-406c-a9e5-66933d921064 184 0.26
a696ccf9-22cb-428b-adee-95c9a97a4581 251 0.36
26c1022a-7a8e-42a2-b7cc-bea6bffa7a6f 253 0.36
...
...
...
It's not clear you came up with 0.26, 0.36 in your desired output - but assuming those are just dummy numbers, to get a running % of total for each group, you could do this:
y['cumsum'] = y.groupby(level=[0,1]).cumsum()
y['running_pct'] = y.groupby(level=[0,1])['cumsum'].transform(lambda x: x / x.iloc[-1])
Should give the right output.
In [398]: y['running_pct'].head()
Out[398]:
app platform uuid
1 0 7e302dcb-ceaf-406c-a9e5-66933d921064 0.727273
a696ccf9-22cb-428b-adee-95c9a97a4581 0.992095
26c1022a-7a8e-42a2-b7cc-bea6bffa7a6f 1.000000
2 0 953a525c-97e0-4c2f-90e0-dfebde3ec20d 0.755332
9c08c860-7a6d-4810-a5c3-f3af2a3fcf66 0.902760
Name: running_pct, dtype: float64
EDIT:
Per the comments, if you're looking to wring out a little more performance, this will be faster as of version 0.14.1
y['cumsum'] = y.groupby(level=[0,1])['minutes'].transform('cumsum')
y['running_pct'] = y['cumsum'] / y.groupby(level=[0,1])['minutes'].transform('sum')
And as #Jeff notes, in 0.15.0 this may be faster yet.
y['running_pct'] = y['cumsum'] / y.groupby(level=[0,1])['minutes'].transform('last')