I have a data frame which is called data.
Also, I have 2057 columns and 197 rows in dataframe group1, I want to know which row is similar to another one.
I made this:
group1=pd.crosstab(data.column1,data.column2)
group1["EsDuplicado?"]=group1.duplicated(subset=group1.columns.difference(['BCP_Nombre_de_la_Matriz__c']),keep=False)
Until now it's working, and I added a new column with value is true(when the row is similar to another one) or false when the row is not similar.
I want To know which rows are similar and know exactly its pair.
Until now my table is like this:
But I would want something like this:
Or maybe this is not necessary, and maybe it's enough if the rows are next to each other, in that way I know which row is similar.
I want something like this, but in this example is only 2 column in my case I have 2057 columns:
find duplicate rows in a pandas dataframe
The answer by #cs95 to the question you linked to can straightforwardly be generalized to any number of columns.
Here's a small example dataset for testing, but I'll make the second code block general, so that it should also work for your DataFrame, as long as its name is group1.
import pandas as pd
group1 = pd.DataFrame({'column 1': [1, 1, 1, 1, 1, 1],
'column 2': [2, 2, 2, 2, 2, 2],
'column 3': [3, 4, 3, 5, 3, 4]})
group1
column 1 column 2 column 3
0 1 2 3
1 1 2 4
2 1 2 3
3 1 2 5
4 1 2 3
5 1 2 4
group1['first such row'] = group1.groupby(list(group1.columns))[group1.columns[0]
].transform('idxmin')
group1
column 1 column 2 column 3 first such row
0 1 2 3 0
1 1 2 4 1
2 1 2 3 0
3 1 2 5 3
4 1 2 3 0
5 1 2 4 1
Related
I have following dataframe:
period symptoms recovery
1 4 2
1 5 2
1 6 2
2 3 1
2 5 2
2 8 4
2 12 6
3 4 2
3 5 2
3 6 3
3 8 5
4 5 2
4 8 4
4 12 6
I'm trying to find the common values of df['period'] groups (1, 2, 3, 4) based on value
of two columns 'symptoms' and 'recovery'
Result should be :
symptoms recovery period
5 2 [1, 2, 3, 4]
8 4 [2, 4]
where each same two columns values has the periods occurrence in a list or column.
I'm I approaching the problem in the wrong way ? Appreciate your help.
I tried to turn each period into dict and loop through to find values but didn't work for me. Also tried to use grouby().apply() but I'm not getting a meaningful data frame.
Tried sorting values based on 3 columns but couldn't get the common ones between each period section.
Last attempt :
df2 = df[['period', 'how_long', 'days_to_ex']].copy()
#s = df.groupby(["period", "symptoms", "recovery"]).size()
s = df.groupby(["symptoms", "recovery"]).size()
You were almost there:
from io import StringIO
import pandas as pd
# setup sample data
data = StringIO("""
period;symptoms;recovery
1;4;2
1;5;2
1;6;2
2;3;1
2;5;2
2;8;4
2;12;6
3;4;2
3;5;2
3;6;3
3;8;5
4;5;2
4;8;4
4;12;6
""")
df = pd.read_csv(data, sep=";")
# collect unique periods
df.groupby(['symptoms','recovery'])[['period']].agg(list).reset_index()
This gives
symptoms recovery period
0 3 1 [2]
1 4 2 [1, 3]
2 5 2 [1, 2, 3, 4]
3 6 2 [1]
4 6 3 [3]
5 8 4 [2, 4]
6 8 5 [3]
7 12 6 [2, 4]
I have a DataFrame like this, where for column2 I need to add 0.004 throughout the column to get a 0 value in row 1 of column 2. Similarly, for column 3 I need to subtract 0.4637 from the entire column to get a 0 value at row 1 column 3. How do I efficiently execute this?
Here is my code -
df2 = pd.DataFrame(np.zeros((df.shape[0], len(df.columns)))).round(0).astype(int)
for (i,j) in zip(range(0, 5999),range(1,len(df.columns))):
if j==1:
df2.values[i,j] = df.values[i,j] + df.values[0,1]
elif j>1:
df2.iloc[i,j] = df.iloc[i,j] - df.iloc[0,j]
print(df2)
Any help would be greatly appreciated. Thank you.
df2 = df - df.iloc[0]
Explanation:
Let's work through an example.
df = pd.DataFrame(np.arange(20).reshape(4, 5))
0
1
2
3
4
0
0
1
2
3
4
1
5
6
7
8
9
2
10
11
12
13
14
3
15
16
17
18
19
df.iloc[0] selects the first row of the dataframe:
0 0
1 1
2 2
3 3
4 4
Name: 0, dtype: int64
This is a Series. The first column printed here is its index (column names of the dataframe), and the second one - the actual values of the first row of the dataframe.
We can convert it to a list to better see its values
df.iloc[0].tolist()
[0, 1, 2, 3, 4]
Then, using broadcasting, we are subtracting each value from the whole column where it has come from.
I have a DataFrame with columns A, B, and C. For each value of A, I would like to select the row with the minimum value in column B.
That is, from this:
df = pd.DataFrame({'A': [1, 1, 1, 2, 2, 2],
'B': [4, 5, 2, 7, 4, 6],
'C': [3, 4, 10, 2, 4, 6]})
A B C
0 1 4 3
1 1 5 4
2 1 2 10
3 2 7 2
4 2 4 4
5 2 6 6
I would like to get:
A B C
0 1 2 10
1 2 4 4
For the moment I am grouping by column A, then creating a value that indicates to me the rows I will keep:
a = data.groupby('A').min()
a['A'] = a.index
to_keep = [str(x[0]) + str(x[1]) for x in a[['A', 'B']].values]
data['id'] = data['A'].astype(str) + data['B'].astype('str')
data[data['id'].isin(to_keep)]
I am sure that there is a much more straightforward way to do this.
I have seen many answers here that use MultiIndex, which I would prefer to avoid.
Thank you for your help.
I feel like you're overthinking this. Just use groupby and idxmin:
df.loc[df.groupby('A').B.idxmin()]
A B C
2 1 2 10
4 2 4 4
df.loc[df.groupby('A').B.idxmin()].reset_index(drop=True)
A B C
0 1 2 10
1 2 4 4
Had a similar situation but with a more complex column heading (e.g. "B val") in which case this is needed:
df.loc[df.groupby('A')['B val'].idxmin()]
The accepted answer (suggesting idxmin) cannot be used with the pipe pattern. A pipe-friendly alternative is to first sort values and then use groupby with DataFrame.head:
data.sort_values('B').groupby('A').apply(DataFrame.head, n=1)
This is possible because by default groupby preserves the order of rows within each group, which is stable and documented behaviour (see pandas.DataFrame.groupby).
This approach has additional benefits:
it can be easily expanded to select n rows with smallest values in specific column
it can break ties by providing another column (as a list) to .sort_values(), e.g.:
data.sort_values(['final_score', 'midterm_score']).groupby('year').apply(DataFrame.head, n=1)
As with other answers, to exactly match the result desired in the question .reset_index(drop=True) is needed, making the final snippet:
df.sort_values('B').groupby('A').apply(DataFrame.head, n=1).reset_index(drop=True)
I found an answer a little bit more wordy, but a lot more efficient:
This is the example dataset:
data = pd.DataFrame({'A': [1,1,1,2,2,2], 'B':[4,5,2,7,4,6], 'C':[3,4,10,2,4,6]})
data
Out:
A B C
0 1 4 3
1 1 5 4
2 1 2 10
3 2 7 2
4 2 4 4
5 2 6 6
First we will get the min values on a Series from a groupby operation:
min_value = data.groupby('A').B.min()
min_value
Out:
A
1 2
2 4
Name: B, dtype: int64
Then, we merge this series result on the original data frame
data = data.merge(min_value, on='A',suffixes=('', '_min'))
data
Out:
A B C B_min
0 1 4 3 2
1 1 5 4 2
2 1 2 10 2
3 2 7 2 4
4 2 4 4 4
5 2 6 6 4
Finally, we get only the lines where B is equal to B_min and drop B_min since we don't need it anymore.
data = data[data.B==data.B_min].drop('B_min', axis=1)
data
Out:
A B C
2 1 2 10
4 2 4 4
I have tested it on very large datasets and this was the only way I could make it work in a reasonable time.
You can sort_values and drop_duplicates:
df.sort_values('B').drop_duplicates('A')
Output:
A B C
2 1 2 10
4 2 4 4
The solution is, as written before ;
df.loc[df.groupby('A')['B'].idxmin()]
If the solution but then if you get an error;
"Passing list-likes to .loc or [] with any missing labels is no longer supported.
The following labels were missing: Float64Index([nan], dtype='float64').
See https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#deprecate-loc-reindex-listlike"
In my case, there were 'NaN' values at column B. So, I used 'dropna()' then it worked.
df.loc[df.groupby('A')['B'].idxmin().dropna()]
You can also boolean indexing the rows where B column is minimal value
out = df[df['B'] == df.groupby('A')['B'].transform('min')]
print(out)
A B C
2 1 2 10
4 2 4 4
I have a large df that looks like:
test = pd.DataFrame({'start': [1, 1, 2, 8, 2000],
'end': [5, 3, 6, 9, 3000]})
start end
0 1 5
1 1 3
2 2 6
3 8 9
4 2000 3000
I want to compare all rows of test and get the combined value ranges:
desired output:
start end
0 1 6
1 8 9
2 2000 3000
I know I can compare within a row, e.g.
test['start'] < test['end]
I'm just not sure of the best/fastest way to compare and combine over millions of rows.
Having a dataframe like this:
month transactions_ids
0 1 [0, 5, 1]
1 2 [7, 4]
2 3 [8, 10, 9, 11]
3 6 [2]
4 9 [3]
For a given transaction_id, I would like to get the month when it took place. Notice that a transaction_id can only be related to one single month.
So for example, given transaction_id = 4, the month would be 2.
I know this can be done in a loop by looking month by month if the transactions_ids related contain the given transaction_id, but I'm wondering if there is any way more efficient than that.
Cheers
The best way in my opinion is to explode your data frame and avoid having python lists in your cells.
df = df.explode('transaction_ids')
which outputs
month transactions_ids
0 1 0
0 1 5
0 1 1
1 2 7
1 2 4
2 3 8
2 3 10
2 3 9
2 3 11
3 6 2
4 9 3
Then, simply
id_to_find = 1 # example
df.loc[df.transactions_ids == id_to_find, 'month']
P.S: be aware of the duplicated indexes that explode outputs. In general, it is better to do explode(...).reset_index(drop=True) for most cases to avoid unwanted behavior.
You can use pandas string methods to find the id in the "list" (it's really just a string as far as pandas is concerned when read in using StringIO):
import pandas as pd
from io import StringIO
data = StringIO("""
month transactions_ids
1 [0,5,1]
2 [7,4]
3 [8,10,9,11]
6 [2]
9 [3]
""")
df = pd.read_csv(data, delim_whitespace=True)
df.loc[df['transactions_ids'].str.contains('4'), 'month']
In case your transactions_ids are real lists, then you can use map to check for membership:
df['transactions_ids'].map(lambda x: 3 in x)