UPDATE: I have edited the question (and code) to make the problem clearer. I use here synthetic data but imagine a large df of floods and a small one of significant floods. I want add a reference to every row (of the large_df) if it is somewhat close to the significant flood.
I have 2 pandas dataframes (1 large and 1 small).
In every iteration I want to create a subset of the small dataframe based on a few conditions that are dependent on each row (of the large df):
import numpy as np
import pandas as pd
import time
SOME_THRESHOLD = 10.5
MUMBER_OF_ROWS = 2e4
large_df = pd.DataFrame(index=np.arange(MUMBER_OF_ROWS), data={'a': np.arange(MUMBER_OF_ROWS)})
small_df = large_df.loc[np.random.randint(0, MUMBER_OF_ROWS, 5)]
large_df['past_index'] = None
count_time = 0
for ind, row in large_df.iterrows():
start = time.time()
# This line takes forever.
df_tmp = small_df[(small_df.index<ind) & (small_df['a']>(row['a']-SOME_THRESHOLD)) & (small_df['a']<(row['a']+SOME_THRESHOLD))]
count_time += time.time()-start
if not df_tmp.empty:
past_index = df_tmp.loc[df_tmp.index.max()]['a']
large_df.loc[ind, 'similar_past_flood_tag'] = f'Similar to the large flood of {past_index}'
print(f'The total time of creating the subset df for 2e4 rows is: {count_time} seconds.')
The line that creates the subset takes a long time to compute:
The total time of creating the subset df for 2e4 rows is: 18.276793956756592 seconds.
This seems to me to be an too long. I have found similar questions but non of the answers seemed to work (e.g query and numpy conditions).
Is there a way to optimize this?
Note: the code does what is expected - just very slow.
While your code is logically correct, building the many boolean arrays and slicing the DataFrame accumulates to some time..
Here are some stats with %timeit:
(small_df.index<ind): ~30μs
(small_df['a']>(row['a']-SOME_THRESHOLD)): ~100μs
(small_df['a']<(row['a']+SOME_THRESHOLD)): ~100μs
After '&'-ing all three: ~500μs
Including the DataFrame slice: ~700μs
That, multiplied by 20K times is indeed 14 seconds.. :)
What you could do is take advantage of numpy's broadcast to compute the boolean matrix more efficiently, and then reconstruct the "valid" DataFrame. See below:
l_ind = np.array(large_df.index)
s_ind = np.array(small_df.index)
l_a = np.array(large_df.a)
s_a = np.array(small_df.a)
arr1 = (l_ind[:, None] < s_ind[None, :])
arr2 = (((l_a[:, None] - SOME_THRESHOLD) < s_a[None, :]) &
(s_a[None, :] < (l_a[:, None] + SOME_THRESHOLD)))
arr = arr1 & arr2
large_valid_inds, small_valid_inds = np.where(arr)
pd.DataFrame({'large_ind': np.take(l_ind, large_valid_inds),
'small_ind': np.take(s_ind, small_valid_inds)})
That gives you the following DF, which if I understood the question properly, is the expected solution:
large_ind
small_ind
0
1621
1631
1
1622
1631
2
1623
1631
3
1624
1631
4
1625
1631
5
1626
1631
6
1627
1631
7
1628
1631
8
1629
1631
9
1630
1631
10
1992
2002
11
1993
2002
12
1994
2002
13
1995
2002
14
1996
2002
15
1997
2002
16
1998
2002
17
1999
2002
18
2000
2002
19
2001
2002
20
8751
8761
21
8752
8761
22
8753
8761
23
8754
8761
24
8755
8761
25
8756
8761
26
8757
8761
27
8758
8761
28
8759
8761
29
8760
8761
30
10516
10526
31
10517
10526
32
10518
10526
33
10519
10526
34
10520
10526
35
10521
10526
36
10522
10526
37
10523
10526
38
10524
10526
39
10525
10526
40
18448
18458
41
18449
18458
42
18450
18458
43
18451
18458
44
18452
18458
45
18453
18458
46
18454
18458
47
18455
18458
48
18456
18458
49
18457
18458
In pandas for loops are much slower than column operations. So changing the calculation to loop over small_df instead of large_df will already give a big improvement:
for ind, row in small_df.iterrows():
df_tmp = large_df[ <some condition> ]
# ... some other processing
Even better for your case is to use a merge rather than a condition on large_df. The problem is your merge is not on equal columns but on approximately equal. To use this approach, you should truncate your column and use that for the merge. Here's a hacky example.
small_df['a_rounded'] = (small_df['a'] / SOME_THRESHOLD / 2).astype(int)
large_df['a_rounded'] = (large_df['a'] / SOME_THRESHOLD / 2).astype(int)
merge_result = small_df.merge(large_df, on='a_rounded')
small_df['a_rounded2'] = ((small_df['a'] + SOME_THRESHOLD) / SOME_THRESHOLD / 2).astype(int)
large_df['a_rounded2'] = ((large_df['a'] + SOME_THRESHOLD) / SOME_THRESHOLD / 2).astype(int)
merge_result2 = small_df.merge(large_df, on='a_rounded2')
total_merge_result = pd.concat([merge_result, merge_result2])
# Now remove duplicates and impose additional filters.
You can impose the additional filters on the result later.
So I learned that I can use DataFrame.groupby without having a MultiIndex to do subsampling/cross-sections.
On the other hand, when I have a MultiIndex on a DataFrame, I still need to use DataFrame.groupby to do sub-sampling/cross-sections.
So what is a MultiIndex good for apart from the quite helpful and pretty display of the hierarchies when printing?
Hierarchical indexing (also referred to as “multi-level” indexing) was introduced in the pandas 0.4 release.
This opens the door to some quite sophisticated data analysis and manipulation, especially for working with higher dimensional data. In essence, it enables you to effectively store and manipulate arbitrarily high dimension data in a 2-dimensional tabular structure (DataFrame), for example.
Imagine constructing a dataframe using MultiIndex like this:-
import pandas as pd
import numpy as np
np.arrays = [['one','one','one','two','two','two'],[1,2,3,1,2,3]]
df = pd.DataFrame(np.random.randn(6,2),index=pd.MultiIndex.from_tuples(list(zip(*np.arrays))),columns=['A','B'])
df # This is the dataframe we have generated
A B
one 1 -0.732470 -0.313871
2 -0.031109 -2.068794
3 1.520652 0.471764
two 1 -0.101713 -1.204458
2 0.958008 -0.455419
3 -0.191702 -0.915983
This df is simply a data structure of two dimensions
df.ndim
2
But we can imagine it, looking at the output, as a 3 dimensional data structure.
one with 1 with data -0.732470 -0.313871.
one with 2 with data -0.031109 -2.068794.
one with 3 with data 1.520652 0.471764.
A.k.a.: "effectively store and manipulate arbitrarily high dimension data in a 2-dimensional tabular structure"
This is not just a "pretty display". It has the benefit of easy retrieval of data since we now have a hierarchal index.
For example.
In [44]: df.ix["one"]
Out[44]:
A B
1 -0.732470 -0.313871
2 -0.031109 -2.068794
3 1.520652 0.471764
will give us a new data frame only for the group of data belonging to "one".
And we can narrow down our data selection further by doing this:-
In [45]: df.ix["one"].ix[1]
Out[45]:
A -0.732470
B -0.313871
Name: 1
And of course, if we want a specific value, here's an example:-
In [46]: df.ix["one"].ix[1]["A"]
Out[46]: -0.73247029752040727
So if we have even more indexes (besides the 2 indexes shown in the example above), we can essentially drill down and select the data set we are really interested in without a need for groupby.
We can even grab a cross-section (either rows or columns) from our dataframe...
By rows:-
In [47]: df.xs('one')
Out[47]:
A B
1 -0.732470 -0.313871
2 -0.031109 -2.068794
3 1.520652 0.471764
By columns:-
In [48]: df.xs('B', axis=1)
Out[48]:
one 1 -0.313871
2 -2.068794
3 0.471764
two 1 -1.204458
2 -0.455419
3 -0.915983
Name: B
Great post by #Calvin Cheng, but thought I'd take a stab at this as well.
When to use a MultiIndex:
When a single column’s value isn’t enough to uniquely identify a row.
When data is logically hierarchical - meaning that it has multiple dimensions or “levels.”
Why (your core question) - at least these are the biggest benefits IMO:
Easy manipulation via stack() and unstack()
Easy math when there are multiple column levels
Syntactic sugar for slicing/filtering
Example:
Dollars Units
Date Store Category Subcategory UPC EAN
2018-07-10 Store 1 Alcohol Liqour 80480280024 154.77 7
Store 2 Alcohol Liqour 80480280024 82.08 4
Store 3 Alcohol Liqour 80480280024 259.38 9
Store 1 Alcohol Liquor 80432400630 477.68 14
674545000001 139.68 4
Store 2 Alcohol Liquor 80432400630 203.88 6
674545000001 377.13 13
Store 3 Alcohol Liquor 80432400630 239.19 7
674545000001 432.32 14
Store 1 Beer Ales 94922755711 65.17 7
702770082018 174.44 14
736920111112 50.70 5
Store 2 Beer Ales 94922755711 129.60 12
702770082018 107.40 10
736920111112 59.65 5
Store 3 Beer Ales 94922755711 154.00 14
702770082018 137.40 10
736920111112 107.88 12
Store 1 Beer Lagers 702770081011 156.24 12
Store 2 Beer Lagers 702770081011 137.06 11
Store 3 Beer Lagers 702770081011 119.52 8
1) If we want to easily compare sales across stores, we can use df.unstack('Store') to line everything up side-by-side:
Dollars Units
Store Store 1 Store 2 Store 3 Store 1 Store 2 Store 3
Date Category Subcategory UPC EAN
2018-07-10 Alcohol Liqour 80480280024 154.77 82.08 259.38 7 4 9
Liquor 80432400630 477.68 203.88 239.19 14 6 7
674545000001 139.68 377.13 432.32 4 13 14
Beer Ales 94922755711 65.17 129.60 154.00 7 12 14
702770082018 174.44 107.40 137.40 14 10 10
736920111112 50.70 59.65 107.88 5 5 12
Lagers 702770081011 156.24 137.06 119.52 12 11 8
2) We can also easily do math on multiple columns. For example, df['Dollars'] / df['Units'] will then divide each store's dollars by its units, for every store without multiple operations:
Store Store 1 Store 2 Store 3
Date Category Subcategory UPC EAN
2018-07-10 Alcohol Liqour 80480280024 22.11 20.52 28.82
Liquor 80432400630 34.12 33.98 34.17
674545000001 34.92 29.01 30.88
Beer Ales 94922755711 9.31 10.80 11.00
702770082018 12.46 10.74 13.74
736920111112 10.14 11.93 8.99
Lagers 702770081011 13.02 12.46 14.94
3) If we then want to filter to just specific rows, instead of using the
df[(df[col1] == val1) and (df[col2] == val2) and (df[col3] == val3)]
format, we can instead .xs or .query (yes these work for regular dfs, but it's not very useful). The syntax would instead be:
df.xs((val1, val2, val3), level=(col1, col2, col3))
More examples can be found in this tutorial notebook I put together.
The alternative to using a multiindex is to store your data using multiple columns of a dataframe. One would expect multiindex to provide a performance boost over naive column storage, but as of Pandas v 1.1.4, that appears not to be the case.
Timinigs
import numpy as np
import pandas as pd
np.random.seed(2020)
inv = pd.DataFrame({
'store_id': np.random.choice(10000, size=10**7),
'product_id': np.random.choice(1000, size=10**7),
'stock': np.random.choice(100, size=10**7),
})
# Create a DataFrame with a multiindex
inv_multi = inv.groupby(['store_id', 'product_id'])[['stock']].agg('sum')
print(inv_multi)
stock
store_id product_id
0 2 48
4 18
5 58
7 149
8 158
... ...
9999 992 132
995 121
996 105
998 99
999 16
[6321869 rows x 1 columns]
# Create a DataFrame without a multiindex
inv_cols = inv_multi.reset_index()
print(inv_cols)
store_id product_id stock
0 0 2 48
1 0 4 18
2 0 5 58
3 0 7 149
4 0 8 158
... ... ... ...
6321864 9999 992 132
6321865 9999 995 121
6321866 9999 996 105
6321867 9999 998 99
6321868 9999 999 16
[6321869 rows x 3 columns]
%%timeit
inv_multi.xs(key=100, level='store_id')
10 loops, best of 3: 20.2 ms per loop
%%timeit
inv_cols.loc[inv_cols.store_id == 100]
The slowest run took 8.79 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 11.5 ms per loop
%%timeit
inv_multi.xs(key=100, level='product_id')
100 loops, best of 3: 9.08 ms per loop
%%timeit
inv_cols.loc[inv_cols.product_id == 100]
100 loops, best of 3: 12.2 ms per loop
%%timeit
inv_multi.xs(key=(100, 100), level=('store_id', 'product_id'))
10 loops, best of 3: 29.8 ms per loop
%%timeit
inv_cols.loc[(inv_cols.store_id == 100) & (inv_cols.product_id == 100)]
10 loops, best of 3: 28.8 ms per loop
Conclusion
The benefits from using a MultiIndex are about syntactic sugar, self-documenting data, and small conveniences from functions like unstack() as mentioned in #ZaxR's answer; Performance is not a benefit, which seems like a real missed opportunity.
Based on the comment on this
answer it seems the
experiment was flawed. Here is my attempt at a correct experiment.
Timings
import pandas as pd
import numpy as np
from timeit import timeit
random_data = np.random.randn(16, 4)
multiindex_lists = [["A", "B", "C", "D"], [1, 2, 3, 4]]
multiindex = pd.MultiIndex.from_product(multiindex_lists)
dfm = pd.DataFrame(random_data, multiindex)
df = dfm.reset_index()
print("dfm:\n", dfm, "\n")
print("df\n", df, "\n")
dfm_selection = dfm.loc[("B", 4), 3]
print("dfm_selection:", dfm_selection, type(dfm_selection))
df_selection = df[(df["level_0"] == "B") & (df["level_1"] == 4)][3].iat[0]
print("df_selection: ", df_selection, type(df_selection), "\n")
print("dfm_selection timeit:",
timeit(lambda: dfm.loc[("B", 4), 3], number=int(1e6)))
print("df_selection timeit: ",
timeit(
lambda: df[(df["level_0"] == "B") & (df["level_1"] == 4)][3].iat[0],
number=int(1e6)))
dfm:
0 1 2 3
A 1 -1.055128 -0.845019 -2.853027 0.521738
2 0.397804 0.385045 -0.121294 -0.696215
3 -0.551836 -0.666953 -0.956578 1.929732
4 -0.154780 1.778150 0.183104 -0.013989
B 1 -0.315476 0.564419 0.492496 -1.052432
2 -0.695300 0.085265 0.701724 -0.974168
3 -0.879915 -0.206499 1.597701 1.294885
4 0.653261 0.279641 -0.800613 1.050241
C 1 1.004199 -1.377520 -0.672913 1.491793
2 -0.453452 0.367264 -0.002362 0.411193
3 2.271958 0.240864 -0.923934 -0.572957
4 0.737893 -0.523488 0.485497 -2.371977
D 1 1.133661 -0.584973 -0.713320 -0.656315
2 -1.173231 -0.490667 0.634677 1.711015
3 -0.050371 -0.175644 0.124797 0.703672
4 1.349595 0.122202 -1.498178 0.013391
df
level_0 level_1 0 1 2 3
0 A 1 -1.055128 -0.845019 -2.853027 0.521738
1 A 2 0.397804 0.385045 -0.121294 -0.696215
2 A 3 -0.551836 -0.666953 -0.956578 1.929732
3 A 4 -0.154780 1.778150 0.183104 -0.013989
4 B 1 -0.315476 0.564419 0.492496 -1.052432
5 B 2 -0.695300 0.085265 0.701724 -0.974168
6 B 3 -0.879915 -0.206499 1.597701 1.294885
7 B 4 0.653261 0.279641 -0.800613 1.050241
8 C 1 1.004199 -1.377520 -0.672913 1.491793
9 C 2 -0.453452 0.367264 -0.002362 0.411193
10 C 3 2.271958 0.240864 -0.923934 -0.572957
11 C 4 0.737893 -0.523488 0.485497 -2.371977
12 D 1 1.133661 -0.584973 -0.713320 -0.656315
13 D 2 -1.173231 -0.490667 0.634677 1.711015
14 D 3 -0.050371 -0.175644 0.124797 0.703672
15 D 4 1.349595 0.122202 -1.498178 0.013391
dfm_selection: 1.0502406808918188 <class 'numpy.float64'>
df_selection: 1.0502406808918188 <class 'numpy.float64'>
dfm_selection timeit: 63.92458086000079
df_selection timeit: 450.4555013199997
Conclusion
MultiIndex single-value retrieval is over 7 times faster than conventional
dataframe single-value retrieval.
The syntax for MultiIndex retrieval is much cleaner.
I have this regular dataframe indexed by 'Date', called ES:
Price Day Hour num_obs med abs_med Ret
Date
2006-01-03 08:30:00 1260.583333 1 8 199 1260.416667 0.166667 0.000364
2006-01-03 08:35:00 1261.291667 1 8 199 1260.697917 0.593750 0.000562
2006-01-03 08:40:00 1261.125000 1 8 199 1260.843750 0.281250 -0.000132
2006-01-03 08:45:00 1260.958333 1 8 199 1260.895833 0.062500 -0.000132
2006-01-03 08:50:00 1261.214286 1 8 199 1260.937500 0.276786 0.000203
I have this other dataframe indexed by the following MultiIndex. The first index goes from 0 to 23 and the second index goes from 0 to 55. In other words we have daily 5 minute increment data.
5min_Ret
0 0 2.235875e-06
5 9.814064e-07
10 -1.453213e-06
15 4.295757e-06
20 5.884896e-07
25 -1.340122e-06
30 9.470660e-06
35 1.178204e-06
40 -1.111621e-05
45 1.159005e-05
50 6.148861e-06
55 1.070586e-05
1 0 1.485287e-05
5 3.018576e-06
10 -1.513273e-05
15 -1.105312e-05
20 3.600874e-06
...
I want to create a column in the original dataframe, ES, that has the appropriate '5min_Ret' at each appropriate hour/5minute combo.
I've tried multiple things: looping over rows, finding some apply function. But nothing has worked so far. I feel like I'm overlooking a simple and Pythonic solution here.
The expected output creates a new column called '5min_ret' to the original dataframe in which each row corresponds to the correct hour/5minute pair from the smaller dataframe containing the 5min_ret
Price Day Hour num_obs med abs_med Ret 5min_ret
Date
2006-01-03 08:30:00 1260.583333 1 8 199 1260.416667 0.166667 0.000364 xxxx
2006-01-03 08:35:00 1261.291667 1 8 199 1260.697917 0.593750 0.000562 xxxx
2006-01-03 08:40:00 1261.125000 1 8 199 1260.843750 0.281250 -0.000132 xxxx
2006-01-03 08:45:00 1260.958333 1 8 199 1260.895833 0.062500 -0.000132 xxxx
2006-01-03 08:50:00 1261.214286 1 8 199 1260.937500 0.276786 0.000203 xxxx
I think one way is to use merge on hour and minute. First create a column 'min' in ES from the datetimeindex such as:
ES['min'] = ES.index.minute
Now you can merge with your multiindex DF containing the column '5min_Ret' that I named df_multi such as:
ES = ES.merge(df_multi.reset_index(), left_on = ['hour','min'],
right_on = ['level_0','level_1'], how='left')
Here you merge on 'hour' and 'min' from ES with 'level_0' and 'level_1', which are created from your multiindex of df_multi when you do reset_index, and on the value of the left df (being ES)
You should get a new column in ES named '5min_Ret' with the value you are looking for. You can drop the colum 'min' if you don't need it anymore by ES = ES.drop('min',axis=1)
i have a pandas dataframe
id no_of_rows
1 2689
2 1515
3 3826
4 814
5 1650
6 2292
7 1867
8 2096
9 1618
10 923
11 766
12 191
i want to divide id's into 5 different bins based on their no. of rows,
such that every bin has approx(equal no of rows)
and assign it as a new column bin
One approach i thought was
df.no_of_rows.sum() = 20247
div_factor = 20247//5 == 4049
if we add 1st and 2nd row its sum = 2689+1515 = 4204 > div_factor.
Therefore assign bin = 1 where id = 1.
Now look for the next ones
id no_of_rows bin
1 2689 1
2 1515 2
3 3826 3
4 814 4
5 1650 4
6 2292 5
7 1867
8 2096
9 1618
10 923
11 766
12 191
But this method proved wrong.
Is there a way to have 5 bins such that every bin has good amount of stores(approximately equal)
You can use an approach based on percentiles.
n_bins = 5
dfa = df.sort_values(by='no_of_rows').cumsum()
df['bin'] = dfa.no_of_rows.apply(lambda x: int(n_bins*x/dfa.no_of_rows.max()))
And then you can check with
df.groupby('bin').sum()
The more records you have the more fair it will be in terms of dispersion.
I'm playing around with Seaborn and Matplotlib and I trying to find the best type of graph to show the correlation between fare values and chance of survival from the titanic dataset.
The Titanic fare column has a lot of different values ranging from 1 to 500 and some of the values are repeated often.
Here is a sample of value_counts:
titanic.fare.value_counts()
8.0500 43
13.0000 42
7.8958 38
7.7500 34
26.0000 31
10.5000 24
7.9250 18
7.7750 16
0.0000 15
7.2292 15
26.5500 15
8.6625 13
7.8542 13
7.2500 13
7.2250 12
16.1000 9
9.5000 9
15.5000 8
24.1500 8
14.5000 7
7.0500 7
52.0000 7
31.2750 7
56.4958 7
69.5500 7
14.4542 7
30.0000 6
39.6875 6
46.9000 6
21.0000 6
.....
91.0792 2
106.4250 2
164.8667 2
Survival column on the other hand has only two values :
>>> titanic.survived.head(10)
271 1
597 0
302 0
633 0
277 0
413 0
674 0
263 0
466 0
A histogram would only show the frequency of fares in certain ranges.
For a scatter plot I would need two variables; having "survived" which has only two values would make for a strange variable.
Is there a way to show the rise of survivability as fare increases clearly through a line graph?
I know there is a correlation as If I sort fare values in ascending order (000-500).
Then do:
>>> titanic.head(50).survived.sum()
5
>>>titanic.tail(50).survived.sum()
37
I see a correlation.
Thanks.
This is what I did to show the correlation between the fare values and the chance of survival:
First, I created a new column Fare Groups, converting fare values to groups of fare ranges, using cut().
df['Fare Groups'] = pd.cut(df.Fare, [0,50,100,150,200,550])
Next, I created a pivot_table().
piv_fare = df.pivot_table(index='Fare Groups', columns='Survived', values = 'Fare', aggfunc='count')
Output:
Survived 0 1
Fare Groups
(0, 50] 484 232
(50, 100] 37 70
(100, 150] 5 19
(150, 200] 3 6
(200, 550] 6 14
Plot:
piv_fare.plot(kind='bar')
It seems, those who had the cheapest tickets (0 to 50) had the lowest chance of survival. In fact, (0 to 50) is the only fare range where the chance to die is higher than the chance to survive. Not just higher, but significantly higher.