There is a some seemingly inconsistent behaviour observed when removing duplicates in pandas.
Problem set up: I have a dataframe with three columns and 3330 timeseries observations as shown below:
data.describe()
Mean Buy Sell
count 3330 3330 3330
Checking if the data contains any duplicates, shows there are duplicate indices.
data.index.duplicated().any()
True
How many duplicates are in the data
data.loc[data.index.duplicated()].count()
Mean 38
Buy 38
Sell 38
The duplicates can be visually inspected too
`data[data.index.duplicated()]`
Dilemma: Clearly, there are duplicates in the data and it seems they are 38 of them per column. However, when I use the DataFrame's drop_duplicates(), it seems more data is dropped than expected.
`data.drop_duplicates().count()`
Mean 3241
Buy 3241
Sell 3241
dtype: int64
`data.count() - data.drop_duplicates().count()`
Mean 89
Buy 89
Sell 89
Any ideas on what is the cause of this disparity or the detail I'm missing would be appreciated. Note: It is possible to have similar entries of data but dates should not be duplicated hence the reasonable way to clean the data is to remove the duplicate days.
If I understand you correctly, you want to keep only the first occurrence (row / record) where there are duplicates in your index?
This will accomplish that.
import pandas as pd
df = pd.DataFrame({'IDX':[1,2,2,2,3,4,5,5,6],
'Mean':[1,2,3,4,5,6,7,8,9]}).set_index('IDX')
df
Mean
IDX
1 1
2 2
2 3
2 4
3 5
4 6
5 7
5 8
6 9
duplicates = df.index.duplicated()
duplicates
array([False, False, True, True, False, False, False, True, False])
keep = duplicates == False
df.loc[keep,:]
Mean
IDX
1 1
2 2
3 5
4 6
5 7
6 9
Related
I've got a df that contains various strings that are associated with unique values. For these unique values, I want to drop the rows that are not equal to a separate list, except for the last row.
Using below, the various string values in Label are associated with Item. So for each unique Item, there could be multiple rows in Label with various strings. I only want to keep the strings that are in label_list, except for the last row.
I'm not sure I can do this another way as the amount of strings not in label_list is too many to account for. The ordering van also vary. So for each unique value in Item, I really only want the last row and whatever rows that are in label_list.
label_list = ['A','B','C','D']
df = pd.DataFrame({
'Item' : [10,10,10,10,10,20,20,20],
'Label' : ['A','X','C','D','Y','A','B','X'],
'Count' : [80.0,80.0,200.0,210.0,260.0,260.0,300.0,310.0],
})
df = df[df['Label'].isin(label_list)]
Intended output:
Item Label Value
0 10 A 80.0
1 10 C 200.0
2 10 D 210.0
3 10 Y 260.0
4 20 A 260.0
5 20 B 300.0
6 20 X 310.0
This comes to mind as a quick and dirty solution:
df = pd.concat([df[df['Label'].isin(label_list)],df.drop_duplicates('Item',keep='last')]).drop_duplicates(keep='first')
We are appending the last row of each Item group, but in case the last row is duplicsted because it is also in label_list we are using drop duplicates for the concatenated outout too.
check if 'Label' isin label_list
check if rows are duplicated
boolean slice the dataframe
isin_ = df['Label'].isin(label_list)
duped = df.duplicated('Item', keep='last')
df[isin_ | ~duped]
Item Label Count
0 10 A 80.0
2 10 C 200.0
3 10 D 210.0
4 10 Y 260.0
5 20 A 260.0
6 20 B 300.0
7 20 X 310.0
I'm trying to label data in the original dataframe, based on multiple boolean conditions. This is easy enough when labeling based on one or two conditions, but as I begin requiring multiple conditions the code becomes difficult to manage. The solution seems to break the code down into copies, but that causes chain errors. Here is one example of the issue...
This is a simplified version of what my data looks like:
df=pd.DataFrame(np.array([['ABC',1,3,3,4], ['std',0,0,2,4],['std',2,1,2,4],['std',4,4,2,4],['std',2,6,2,6]]), columns=['Note', 'Na','Mg','Si','S'])
df
Note Na Mg Si S
0 ABC 1 3 3 4
1 std 0 0 2 4
2 std 2 1 2 4
3 std 4 4 2 4
4 std 2 6 2 6
A standard (std) is located throughout the dataframe. I would like to create a label when the instrument fails. This occurs in the data when:
String condition met (Note = standard/std)
Na>0 & Mg>0
Doesn't fall outside of a calculated range for more than 2 elements.
For requirement 3 - Here is an example of a range:
maxMin=pd.DataFrame(np.array([['Max',3,3,3,7], ['Min',1,1,2,2]]), columns=['Note', 'Na','Mg','Si','S'])
maxMin
Note Na Mg Si S
0 Max 3 3 3 7
1 Min 1 1 2 2
Calculating out of bound standard:
elements=['Na','Mg','Si','S']
std=df[(df['Note'].str.contains('std|standard'))&(df['Na']>0)&(df['Mg'])
df.loc[(std[elements].lt(maxMin.loc[1, :])|std[elements].gt(maxMin.loc[0, :]).select_dtypes(include=['bool'])).sum(axis=1)>2]
Note Na Mg Si S
3 std 4 4 2 4
Now, I would like to label this datapoint within the original dataframe. Desired result:
Note Na Mg Si S Error
0 ABC 1 3 3 4 False
1 std 0 0 2 4 False
2 std 2 1 2 4 False
3 std 4 4 2 4 True
4 std 2 6 2 6 False
I've tried things like:
df['Error'].loc[std.loc[(std[elements].lt(maxMin.loc[1, :])|std[elements].gt(mMmaxMinloc[0, :]).select_dtypes(include=['bool'])).sum(axis=1)>5].index.values.copy()]=True
That unfortunately causes a chain error.
How would you accomplish this without creating a chain error? Most books/tutorial revolve around creating one long expression, but as I dive deeper, I feel there might be a simpler solution. Any input would be appreciated
I figured it out a solution that works for me.
The solution was to use .index.value to create an array of the index that passed the bool conditions. That array can be used to pass edit the original dataframe.
##These two conditions can probably be combined
condition1=df[(df['Note'].str.contains('std|standard'))&(df['Na']>.01)&(df['Mg']>.01)]
##where condition1 is greater/less than the bounds of the known value.
##provides array where condition is true
OutofBounds=condition1.loc[(condition1[elements].lt(maxMin.loc[1, :])|condition1[elements].gt(maxMin.loc[0, :]).select_dtypes(include=['bool'])).sum(axis=1)>5].index.values
OutofBounds
out:array([ 3], dtype=int64)
Now I can pass the array into the original dataframe:
df.loc[OutofBounds, 'Error']=True
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'm struggling with a bit of a complex (to me) lookup-type problem.
I have a dataframe df1 that looks like this:
Grade Value
0 A 16
1 B 12
2 C 5
And another dataframe (df2) where the values in one of the columns('Grade') from df1 forms the index:
Tier 1 Tier 2 Tier 3
A 20 17 10
B 16 11 3
C 7 6 2
I've been trying to write a bit of code that for each row in df1, look ups the row corresponding with 'Grade' in df2, finds the smallest value in df2 greater than 'Value', and returns the name of that column.
E.g. for the second row of df1, it looks up the row with index 'B' in df2: 16 is the smallest value greater than 12, so it returns 'Tier 1'. Ideal output would be:
Grade Value Tier
0 A 16 Tier 2
1 B 12 Tier 1
2 C 5 Tier 2
My novice, downloaded-Python-last-week attempt so far has been as follows, which is throwing up all manner of errors and doesn't even try returning the column name. Sorry also about the micro-ness of the question: any help appreciated!
for i, row in input_df1.iterrows():
Tier = np.argmin(df1['Value']<df2.loc[row,0:df2.shape[1]])
df2.loc[df1.Grade].eq(df1.Value, 0).idxmax(1)
Consider following dataframe which has columns with same name (Apparently this does happens, currently I have a dataset like this! :( )
>>> df = pd.DataFrame({"a":range(10,15),"b":range(5,10)})
>>> df.rename(columns={"b":"a"},inplace=True)
df
a a
0 10 5
1 11 6
2 12 7
3 13 8
4 14 9
>>> df.columns
Index(['a', 'a'], dtype='object')
I would expect that when dropping by index , only the column with the respective index would be gone, but apparently this is not the case.
>>> df.drop(df.columns[-1],1)
0
1
2
3
4
Is there a way to get rid of columns with duplicated column names?
EDIT: I choose missleading values for the first column, fixed now
EDIT2: the expected outcome is
a
0 10
1 11
2 12
3 13
4 14
Actually just do this:
In [183]:
df.ix[:,~df.columns.duplicated()]
Out[183]:
a
0 0
1 1
2 2
3 3
4 4
So this index all rows and then uses the column mask generated from duplicated and invert the mask using ~
The output from duplicated:
In [184]:
df.columns.duplicated()
Out[184]:
array([False, True], dtype=bool)
UPDATE
As .ix is deprecated (since v0.20.1) you should do any of the following:
df.iloc[:,~df.columns.duplicated()]
or
df.loc[:,~df.columns.duplicated()]
Thanks to #DavideFiocco for alerting me