I have a pandas dataframe:
df = pd.DataFrame({'one' : [1, 2, 3, 4] ,'two' : [5, 6, 7, 8]})
one two
0 1 5
1 2 6
2 3 7
3 4 8
Column "one" and column "two" together comprise (x,y) coordinates
Lets say I have a list of coordinates: c = [(1,5), (2,6), (20,5)]
Is there an elegant way of obtaining the rows in df
with matching coordinates? In this case, given c, the matching rows would be 0 and 1
Related question: Using pandas to select rows using two different columns from dataframe?
And: Selecting rows from pandas DataFrame using two columns
This approaching using pd.merge should perform better than the iterative solutions.
import pandas as pd
df = pd.DataFrame({"one" : [1, 2, 3, 4] ,"two" : [5, 6, 7, 8]})
c = [(1, 5), (2, 6), (20, 5)]
df2 = pd.DataFrame(c, columns=["one", "two"])
pd.merge(df, df2, on=["one", "two"], how="inner")
one two
0 1 5
1 2 6
You can use
>>> set.union(*(set(df.index[(df.one == i) & (df.two == j)]) for i, j in c))
{0, 1}
Related
I am stuck with an issue on a massive pandas table. I would like to get a boolean to check the cross of 2 series.
df = pd.DataFrame({'A': [1, 2, 3, 4],
'B': [10, 1, 2, 8]})
I would like to add one column in my array to get a result like this one
df = pd.DataFrame({'A': [1, 2, 3, 4],
'B': [10, 1, 2, 8],
'C': [0, -1, 0, 1]
})
So basically to get
0 when there is no cross between series B and A
-1 when table B crosses down table A
1 when table B crosses up table A
I need to do vector calculation because my real table is like more than one million rows.
Thank you
You can compute the relative position of the 2 columns with lt, then convert to integer and compute the diff:
m = df['A'].lt(df['B'])
df['C'] = m.astype(int).diff().fillna(0, downcast='infer')
output:
A B C
0 1 10 0
1 2 1 -1
2 3 2 0
3 4 8 1
visual of A/B:
I'm trying to extract a dataframe which only shows duplicates with e.g 3 or more duplicates in a column. For example:
df = pd.DataFrame({
'one': pd.Series(['Berlin', 'Berlin', 'Tokyo', 'Stockholm','Berlin','Stockholm','Amsterdam']),
'two': pd.Series([1, 2, 3, 4, 5, 6, 7]),
'three': pd.Series([8, 9, 10, 11, 12])
})
Expected output:
one two three
0 Berlin 1 8
The extraction should only show the row of the first duplicate.
You could do it like this:
rows = df.groupby('one').filter(lambda group: group.shape[0] >= 3).groupby('one').first()
Output:
>>> rows
two three
one
Amsterdam 7 1.0
Berlin 1 8.0
It works with multiple groups of 3+ duplicates, too. I tested it.
I have a dataframe that has duplicated time indices and I would like to get the mean across all for the previous 2 days (I do not want to drop any observations; they are all information that I need). I've checked pandas documentation and read previous posts on Stackoverflow (such as Apply rolling mean function on data frames with duplicated indices in pandas), but could not find a solution. Here's an example of how my data frame look like and the output I'm looking for. Thank you in advance.
data:
import pandas as pd
df = pd.DataFrame({'id': [1,1,1,2,3,3,4,4,4],'t': [1, 2, 3, 2, 1, 2, 2, 3, 4],'v1':[1, 2, 3, 4, 5, 6, 7, 8, 9]})
output:
t
v2
1
-
2
-
3
4.167
4
5
5
6.667
A rough proposal to concatenate 2 copies of the input frame in which values in 't' are replaced respectively by values of 't+1' and 't+2'. This way, the meaning of the column 't' becomes "the target day".
Setup:
import pandas as pd
df = pd.DataFrame({'id': [1,1,1,2,3,3,4,4,4],
't': [1, 2, 3, 2, 1, 2, 2, 3, 4],
'v1':[1, 2, 3, 4, 5, 6, 7, 8, 9]})
Implementation:
len = df.shape[0]
incr = pd.DataFrame({'id': [0]*len, 't': [1]*len, 'v1':[0]*len}) # +1 in 't'
df2 = pd.concat([df + incr, df + incr + incr]).groupby('t').mean()
df2 = df2[1:-1] # Drop the days that have no full values for the 2 previous days
df2 = df2.rename(columns={'v1': 'v2'}).drop('id', axis=1)
Output:
v2
t
3 4.166667
4 5.000000
5 6.666667
Thank you for all the help. I ended up using groupby + rolling (2 Day), and then drop duplicates (keep the last observation).
I have the following Pandas dataframe in Python:
import pandas as pd
d = {'col1': [1, 2, 3, 4, 5], 'col2': [6, 7, 8, 9, 10]}
df = pd.DataFrame(data=d)
df.index=['A', 'B', 'C', 'D', 'E']
df
which gives the following output:
col1 col2
A 1 6
B 2 7
C 3 8
D 4 9
E 5 10
I need to write a function (say the name will be getNrRows(fromIndex) ) that will take an index value as input and will return the number of rows between that given index and the last index of the dataframe.
For instance:
nrRows = getNrRows("C")
print(nrRows)
> 2
Because it takes 2 steps (rows) from the index C to the index E.
How can I write such a function in the most elegant way?
The simplest way might be
len(df[row_index:]) - 1
For your information we have built-in function get_indexer_for
len(df)-df.index.get_indexer_for(['C'])-1
Out[179]: array([2], dtype=int64)
I have a dataframe below:
import pandas as pd
d = {'id': [1, 2, 3, 4, 4, 6, 1, 8, 9], 'cluster': [7, 2, 3, 3, 3, 6, 7, 8, 8]}
df = pd.DataFrame(data=d)
df = df.sort_values('cluster')
I want to keep ALL the rows
if there is the same cluster but different id AND keep every row from that cluster
even if it is the same id since there was a different id AT LEAST once within that cluster.
The code I have been using to achieve this is the following below, BUT, the only problem
with this is it drops too many rows for what I am looking for.
df = (df.assign(counts=df.count(axis=1))
.sort_values(['id', 'counts'])
.drop_duplicates(['id','cluster'], keep='last')
.drop('counts', axis=1))
The output dataframe I am expecting that the code above does not do
would drop rows at
dataframe index 1, 5, 0, and 6 but leave dataframe indexes 2, 3, 4, 7, and 8. Essentially
resulting in what the code below produces:
df = df.loc[[2, 3, 4, 7, 8]]
I have looked at many deduplication pandas posts on stack overflow but have yet to find this
scenario. Any help would be greatly appreciated.
I think we can do this with a single boolean. using .groupby().nunique()
con1 = df.groupby('cluster')['id'].nunique() > 1
#of these we only want the True indexes.
cluster
2 False
3 True
6 False
7 False
8 True
df.loc[(df['cluster'].isin(con1[con1].index))]
id cluster
2 3 3
3 4 3
4 4 3
7 8 8
8 9 8