Parsing out indeces and values from pandas multi index dataframe - python

I have a dataframe in a similar format to this:
+--------+--------+----------+------+------+------+------+
| | | | | day1 | day2 | day3 |
+--------+--------+----------+------+------+------+------+
| id_one | id_two | id_three | date | | | |
| 18273 | 50 | 1 | 3 | 9 | 11 | 3 |
| | | | 4 | 26 | 27 | 68 |
| | | | 5 | 92 | 25 | 4 |
| | | | 6 | 60 | 72 | 83 |
| | 60 | 2 | 5 | 69 | 93 | 84 |
| | | | 6 | 69 | 30 | 12 |
| | | | 7 | 65 | 65 | 59 |
| | | | 8 | 57 | 88 | 59 |
| | 70 | 3 | 5 | 22 | 95 | 7 |
| | | | 6 | 40 | 24 | 20 |
| | | | 7 | 73 | 81 | 57 |
| | | | 8 | 43 | 8 | 66 |
+--------+--------+----------+------+------+------+------+
I am trying to create tuple that contains id_one, id_two and the values that each grouping contains.
To test this, I am simply trying to print the ids and values like this:
for id_two, data in df.head(100).groupby(level='id_two'):
print id_two, data.values.ravel()
Which gives me the id_two and the data exactly as it should.
I am running into problems when I try and incorporate id_one. I tried this, but was met with an error ValueError: need more than 2 values to unpack
for id_one, id_two, data in df.head(100).groupby(level='id_two'):
print id_one, id_two, data.values.ravel()
How can I print id_one, id_two and the data?


You can pass a list of columns into the level parameter:
df.head.groupby(level=['id_one', 'id_two'])

Related

How to get the column values of a Dataframe into another dataframe as a new column after matching the values in columns that both dataframes have?

I'm trying to create a new column in a DataFrame and storing it with values stored in a different dataframe by first comparing the values of columns that both dataframes have. For example:
df1 >>>
| name | team | week | dates | interceptions | pass_yds | rating |
| ---- | ---- | -----| ---------- | ------------- | --------- | -------- |
| maho | KC | 1 | 2020-09-10 | 0 | 300 | 105 |
| went | PHI | 1 | 2020-09-13 | 2 | 225 | 74 |
| lock | DEN | 1 | 2020-09-14 | 0 | 150 | 89 |
| dris | DEN | 2 | 2020-09-20 | 1 | 220 | 95 |
| went | PHI | 2 | 2020-09-20 | 2 | 250 | 64 |
| maho | KC | 2 | 2020-09-21 | 1 | 245 | 101 |
df2 >>>
| name | team | week | catches | rec_yds | rec_tds |
| ---- | ---- | -----| ------- | ------- | ------- |
| ertz | PHI | 1 | 5 | 58 | 1 |
| fant | DEN | 2 | 6 | 79 | 0 |
| kelc | KC | 2 | 8 | 105 | 1 |
| fant | DEN | 1 | 3 | 29 | 0 |
| kelc | KC | 1 | 6 | 71 | 1 |
| ertz | PHI | 2 | 7 | 91 | 2 |
| goed | PHI | 2 | 2 | 15 | 0 |
I want to create a dates column in df2 with the values of the dates stored in the dates column in df1 after matching the teams and the weeks columns. After the matching, df2 in this example should look something like this:
df2 >>>
| name | team | week | catches | rec_yds | rec_tds | dates |
| ---- | ---- | -----| ------- | ------- | ------- | ---------- |
| ertz | PHI | 1 | 5 | 58 | 1 | 2020-09-13 |
| fant | DEN | 2 | 6 | 79 | 0 | 2020-09-20 |
| kelc | KC | 2 | 8 | 105 | 1 | 2020-09-20 |
| fant | DEN | 1 | 3 | 29 | 0 | 2020-09-14 |
| kelc | KC | 1 | 6 | 71 | 1 | 2020-09-10 |
| ertz | PHI | 2 | 7 | 91 | 2 | 2020-09-20 |
| goed | PHI | 2 | 2 | 15 | 0 | 2020-09-20 |
I'm looking for an optimal solution. I've already tried nested for loops and comparing the week and team columns from both dataframes together but that hasn't worked. At this point I'm all out of ideas. Please help!
Disclaimer: The actual DataFrames I'm working with are a lot larger. They have a lot more rows, columns, and values (i.e. a lot more teams in the team columns, a lot more dates in the dates columns, and a lot more weeks in the week columns)

Binning Pandas value_counts

I have a Pandas Series produced by df.column.value_counts().sort_index().
| N Months | Count |
|------|------|
| 0 | 15 |
| 1 | 9 |
| 2 | 78 |
| 3 | 151 |
| 4 | 412 |
| 5 | 181 |
| 6 | 543 |
| 7 | 175 |
| 8 | 409 |
| 9 | 594 |
| 10 | 137 |
| 11 | 202 |
| 12 | 170 |
| 13 | 446 |
| 14 | 29 |
| 15 | 39 |
| 16 | 44 |
| 17 | 253 |
| 18 | 17 |
| 19 | 34 |
| 20 | 18 |
| 21 | 37 |
| 22 | 147 |
| 23 | 12 |
| 24 | 31 |
| 25 | 15 |
| 26 | 117 |
| 27 | 8 |
| 28 | 38 |
| 29 | 23 |
| 30 | 198 |
| 31 | 29 |
| 32 | 122 |
| 33 | 50 |
| 34 | 60 |
| 35 | 357 |
| 36 | 329 |
| 37 | 457 |
| 38 | 609 |
| 39 | 4744 |
| 40 | 1120 |
| 41 | 591 |
| 42 | 328 |
| 43 | 148 |
| 44 | 46 |
| 45 | 10 |
| 46 | 1 |
| 47 | 1 |
| 48 | 7 |
| 50 | 2 |
my desired output is
| bin | Total |
|-------|--------|
| 0-13 | 3522 |
| 14-26 | 793 |
| 27-50 | 9278 |
I tried df.column.value_counts(bins=3).sort_index() but got
| bin | Total |
|---------------------------------|-------|
| (-0.051000000000000004, 16.667] | 3634 |
| (16.667, 33.333] | 1149 |
| (33.333, 50.0] | 8810 |
I can get the correct result with
a = df.column.value_counts().sort_index()[:14].sum()
b = df.column.value_counts().sort_index()[14:27].sum()
c = df.column.value_counts().sort_index()[28:].sum()
print(a, b, c)
Output: 3522 793 9270
But I am wondering if there is a pandas method that can do what I want. Any advice is very welcome. :-)

You can use pd.cut:
pd.cut(df['N Months'], [0,13, 26, 50], include_lowest=True).value_counts()
Update you should be able to pass custom bin to value_counts:
df['N Months'].value_counts(bins = [0,13, 26, 50])
Output:
N Months
(-0.001, 13.0] 3522
(13.0, 26.0] 793
(26.0, 50.0] 9278
Name: Count, dtype: int64

Filter all rows from groupby object

I have a dataframe like below
+-----------+------------+---------------+------+-----+-------+
| InvoiceNo | CategoryNo | Invoice Value | Item | Qty | Price |
+-----------+------------+---------------+------+-----+-------+
| 1 | 1 | 77 | 128 | 1 | 10 |
| 1 | 1 | 77 | 101 | 1 | 11 |
| 1 | 2 | 77 | 105 | 3 | 12 |
| 1 | 3 | 77 | 129 | 2 | 10 |
| 2 | 1 | 21 | 145 | 1 | 9 |
| 2 | 2 | 21 | 130 | 1 | 12 |
+-----------+------------+---------------+------+-----+-------+
I want to filter the entire group, if any of the items in the list item_list = [128,129,130] is present in that group, after grouping by 'InvoiceNo' &'CategoryNo'.
My desired out put is as below
+-----------+------------+---------------+------+-----+-------+
| InvoiceNo | CategoryNo | Invoice Value | Item | Qty | Price |
+-----------+------------+---------------+------+-----+-------+
| 1 | 1 | 77 | 128 | 1 | 10 |
| 1 | 1 | 77 | 101 | 1 | 11 |
| 1 | 3 | 77 | 129 | 2 | 10 |
| 2 | 2 | 21 | 130 | 1 | 12 |
+-----------+------------+---------------+------+-----+-------+
I know how to filter a dataframe using isin(). But, not sure how to do it with groupby()
so far i have tried below
import pandas as pd
df = pd.read_csv('data.csv')
item_list = [128,129,130]
df.groupby(['InvoiceNo','CategoryNo'])['Item'].isin(item_list)
but nothing happens. please guide me how to solve this issue.

You can do something like this:
s = (df['Item'].isin(item_list)
.groupby([df['InvoiceNo'], df['CategoryNo']])
.transform('any')
)
df[s]

pandas column shift with day 0 value as 0

I've got a pandas dataframe(pivoted) like customer_name, current_date, current_day_count
+----------+--------------+-------------------+
| customer | current_date | current_day_count |
+----------+--------------+-------------------+
| Mark | 2018_02_06 | 15 |
| | 2018_02_09 | 42 |
| | 2018_02_12 | 33 |
| | 2018_02_21 | 82 |
| | 2018_02_27 | 72 |
| Bob | 2018_02_02 | 76 |
| | 2018_02_23 | 11 |
| | 2018_03_04 | 59 |
| | 2018_03_13 | 68 |
| Shawn | 2018_02_11 | 71 |
| | 2018_02_15 | 39 |
| | 2018_02_18 | 65 |
| | 2018_02_24 | 38 |
+----------+--------------+-------------------+
Now, I want another new column with previous_day_counts for each customer but the first day of the customer's previous day value should be 0 something like this customer, current_date, current_day_count, previous_day_count (with first day value as 0)
+----------+--------------+-------------------+--------------------+
| customer | current_date | current_day_count | previous_day_count |
+----------+--------------+-------------------+--------------------+
| Mark | 2018_02_06 | 15 | 0 |
| | 2018_02_09 | 42 | 33 |
| | 2018_02_12 | 33 | 82 |
| | 2018_02_21 | 82 | 72 |
| | 2018_02_27 | 72 | 0 |
| Bob | 2018_02_02 | 76 | 0 |
| | 2018_02_23 | 11 | 59 |
| | 2018_03_04 | 59 | 68 |
| | 2018_03_13 | 68 | 0 |
| Shawn | 2018_02_11 | 71 | 0 |
| | 2018_02_15 | 39 | 65 |
| | 2018_02_18 | 65 | 38 |
| | 2018_02_24 | 38 | 0 |
+----------+--------------+-------------------+--------------------+

Try this:
import pandas as pd
import numpy as np
df = pd.DataFrame({'name': ['Mark','Mark','Mark','Mark','Bob','Bob','Bob','Bob'], 'current_day_count': [18,28,29,10,19,92,7,43]})
df['previous_day_count'] = df.groupby('name')['current_day_count'].shift(-1)
df.loc[df.groupby('name',as_index=False).head(1).index,'previous_day_count'] = np.nan
df['previous_day_count'].fillna(0, inplace=True)

Numpy version of finding the highest and lowest value locations within an interval of another column?

Given the following numpy array. How can I find the highest and lowest value locations of column 0 within the interval on column 1 using numpy?
import numpy as np
data = np.array([
[1879.289,np.nan],[1879.281,np.nan],[1879.292,1],[1879.295,1],[1879.481,1],[1879.294,1],[1879.268,1],
[1879.293,1],[1879.277,1],[1879.285,1],[1879.464,1],[1879.475,1],[1879.971,1],[1879.779,1],
[1879.986,1],[1880.791,1],[1880.29,1],[1879.253,np.nan],[1878.268,np.nan],[1875.73,1],[1876.792,1],
[1875.977,1],[1876.408,1],[1877.159,1],[1877.187,1],[1883.164,1],[1883.171,1],[1883.495,1],
[1883.962,1],[1885.158,1],[1885.974,1],[1886.479,np.nan],[1885.969,np.nan],[1884.693,1],[1884.977,1],
[1884.967,1],[1884.691,1],[1886.171,1],[1886.166,np.nan],[1884.476,np.nan],[1884.66,1],[1882.962,1],
[1881.496,1],[1871.163,1],[1874.985,1],[1874.979,1],[1871.173,np.nan],[1871.973,np.nan],[1871.682,np.nan],
[1872.476,np.nan],[1882.361,1],[1880.869,1],[1882.165,1],[1881.857,1],[1880.375,1],[1880.66,1],
[1880.891,1],[1880.377,1],[1881.663,1],[1881.66,1],[1877.888,1],[1875.69,1],[1875.161,1],
[1876.697,np.nan],[1876.671,np.nan],[1879.666,np.nan],[1877.182,np.nan],[1878.898,1],[1878.668,1],[1878.871,1],
[1878.882,1],[1879.173,1],[1878.887,1],[1878.68,1],[1878.872,1],[1878.677,1],[1877.877,1],
[1877.669,1],[1877.69,1],[1877.684,1],[1877.68,1],[1877.885,1],[1877.863,1],[1877.674,1],
[1877.676,1],[1877.687,1],[1878.367,1],[1878.179,1],[1877.696,1],[1877.665,1],[1877.667,np.nan],
[1878.678,np.nan],[1878.661,1],[1878.171,1],[1877.371,1],[1877.359,1],[1878.381,1],[1875.185,1],
[1875.367,np.nan],[1865.492,np.nan],[1865.495,1],[1866.995,1],[1866.672,1],[1867.465,1],[1867.663,1],
[1867.186,1],[1867.687,1],[1867.459,1],[1867.168,1],[1869.689,1],[1869.693,1],[1871.676,1],
[1873.174,1],[1873.691,np.nan],[1873.685,np.nan]
])
In the third column below you can see where the max and min is for each interval.
+-------+----------+-----------+---------+
| index | Value | Intervals | Min/Max |
+-------+----------+-----------+---------+
| 0 | 1879.289 | np.nan | |
| 1 | 1879.281 | np.nan | |
| 2 | 1879.292 | 1 | |
| 3 | 1879.295 | 1 | |
| 4 | 1879.481 | 1 | |
| 5 | 1879.294 | 1 | |
| 6 | 1879.268 | 1 | -1 | min
| 7 | 1879.293 | 1 | |
| 8 | 1879.277 | 1 | |
| 9 | 1879.285 | 1 | |
| 10 | 1879.464 | 1 | |
| 11 | 1879.475 | 1 | |
| 12 | 1879.971 | 1 | |
| 13 | 1879.779 | 1 | |
| 17 | 1879.986 | 1 | |
| 18 | 1880.791 | 1 | 1 | max
| 19 | 1880.29 | 1 | |
| 55 | 1879.253 | np.nan | |
| 56 | 1878.268 | np.nan | |
| 57 | 1875.73 | 1 | -1 |min
| 58 | 1876.792 | 1 | |
| 59 | 1875.977 | 1 | |
| 60 | 1876.408 | 1 | |
| 61 | 1877.159 | 1 | |
| 62 | 1877.187 | 1 | |
| 63 | 1883.164 | 1 | |
| 64 | 1883.171 | 1 | |
| 65 | 1883.495 | 1 | |
| 66 | 1883.962 | 1 | |
| 67 | 1885.158 | 1 | |
| 68 | 1885.974 | 1 | 1 | max
| 69 | 1886.479 | np.nan | |
| 70 | 1885.969 | np.nan | |
| 71 | 1884.693 | 1 | |
| 72 | 1884.977 | 1 | |
| 73 | 1884.967 | 1 | |
| 74 | 1884.691 | 1 | -1 | min
| 75 | 1886.171 | 1 | 1 | max
| 76 | 1886.166 | np.nan | |
| 77 | 1884.476 | np.nan | |
| 78 | 1884.66 | 1 | 1 | max
| 79 | 1882.962 | 1 | |
| 80 | 1881.496 | 1 | |
| 81 | 1871.163 | 1 | -1 | min
| 82 | 1874.985 | 1 | |
| 83 | 1874.979 | 1 | |
| 84 | 1871.173 | np.nan | |
| 85 | 1871.973 | np.nan | |
| 86 | 1871.682 | np.nan | |
| 87 | 1872.476 | np.nan | |
| 88 | 1882.361 | 1 | 1 | max
| 89 | 1880.869 | 1 | |
| 90 | 1882.165 | 1 | |
| 91 | 1881.857 | 1 | |
| 92 | 1880.375 | 1 | |
| 93 | 1880.66 | 1 | |
| 94 | 1880.891 | 1 | |
| 95 | 1880.377 | 1 | |
| 96 | 1881.663 | 1 | |
| 97 | 1881.66 | 1 | |
| 98 | 1877.888 | 1 | |
| 99 | 1875.69 | 1 | |
| 100 | 1875.161 | 1 | -1 | min
| 101 | 1876.697 | np.nan | |
| 102 | 1876.671 | np.nan | |
| 103 | 1879.666 | np.nan | |
| 111 | 1877.182 | np.nan | |
| 112 | 1878.898 | 1 | |
| 113 | 1878.668 | 1 | |
| 114 | 1878.871 | 1 | |
| 115 | 1878.882 | 1 | |
| 116 | 1879.173 | 1 | 1 | max
| 117 | 1878.887 | 1 | |
| 118 | 1878.68 | 1 | |
| 119 | 1878.872 | 1 | |
| 120 | 1878.677 | 1 | |
| 121 | 1877.877 | 1 | |
| 122 | 1877.669 | 1 | |
| 123 | 1877.69 | 1 | |
| 124 | 1877.684 | 1 | |
| 125 | 1877.68 | 1 | |
| 126 | 1877.885 | 1 | |
| 127 | 1877.863 | 1 | |
| 128 | 1877.674 | 1 | |
| 129 | 1877.676 | 1 | |
| 130 | 1877.687 | 1 | |
| 131 | 1878.367 | 1 | |
| 132 | 1878.179 | 1 | |
| 133 | 1877.696 | 1 | |
| 134 | 1877.665 | 1 | -1 | min
| 135 | 1877.667 | np.nan | |
| 136 | 1878.678 | np.nan | |
| 137 | 1878.661 | 1 | 1 | max
| 138 | 1878.171 | 1 | |
| 139 | 1877.371 | 1 | |
| 140 | 1877.359 | 1 | |
| 141 | 1878.381 | 1 | |
| 142 | 1875.185 | 1 | -1 | min
| 143 | 1875.367 | np.nan | |
| 144 | 1865.492 | np.nan | |
| 145 | 1865.495 | 1 | -1 | min
| 146 | 1866.995 | 1 | |
| 147 | 1866.672 | 1 | |
| 148 | 1867.465 | 1 | |
| 149 | 1867.663 | 1 | |
| 150 | 1867.186 | 1 | |
| 151 | 1867.687 | 1 | |
| 152 | 1867.459 | 1 | |
| 153 | 1867.168 | 1 | |
| 154 | 1869.689 | 1 | |
| 155 | 1869.693 | 1 | |
| 156 | 1871.676 | 1 | |
| 157 | 1873.174 | 1 | 1 | max
| 158 | 1873.691 | np.nan | |
| 159 | 1873.685 | np.nan | |
+-------+----------+-----------+---------+
I must specify upfront that this question has already been answered here with a pandas solution. The solution performs reasonable at about 300 seconds for a table of around 1 million rows. But after some more testing, I see that if the table is over 3 million rows, the execution time increases dramatically to over 2500 seconds and even more. This is obviously too long for such a simple task. How would the same problem be solved with numpy?

Here's one NumPy approach -
mask = ~np.isnan(data[:,1])
s0 = np.flatnonzero(mask[1:] > mask[:-1])+1
s1 = np.flatnonzero(mask[1:] < mask[:-1])+1
lens = s1 - s0
tags = np.repeat(np.arange(len(lens)), lens)
idx = np.lexsort((data[mask,0], tags))
starts = np.r_[0,lens.cumsum()]
offsets = np.r_[s0[0], s0[1:] - s1[:-1]]
offsets_cumsum = offsets.cumsum()
min_ids = idx[starts[:-1]] + offsets_cumsum
max_ids = idx[starts[1:]-1] + offsets_cumsum
out = np.full(data.shape[0], np.nan)
out[min_ids] = -1
out[max_ids] = 1

So this is a bit of a cheat since it uses scipy:
import numpy as np
from scipy import ndimage
markers = np.isnan(data[:, 1])
groups = np.cumsum(markers)
mins, max, min_idx, max_idx = ndimage.measurements.extrema(
data[:, 0], labels=groups, index=range(2, groups.max(), 2))

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