I have a dataframe of locations:
df = X Y
1 1
2 1
2 1
2 2
3 3
5 5
5.5 5.5
I want to add a columns, with the distance to the previous point:
So it will be:
df = X Y Distance
1 1 0
2 1 1
2 1 0
2 2 1
3 3 2
5 5 2
5.5 5.5 1
What is the best way to do so?
You can use the pd.Series.diff method.
For instance, to compute the eulerian distance, using also np.sqrt, you would do like this:
import numpy as np
df["Distance"] = np.sqrt(df.X.diff()**2 + df.Y.diff()**2)
Related
I'm using groupby on a pandas dataframe to drop all rows that don't have the minimum of a specific column. Something like this:
df1 = df.groupby("item", as_index=False)["diff"].min()
However, if I have more than those two columns, the other columns (e.g. otherstuff in my example) get dropped. Can I keep those columns using groupby, or am I going to have to find a different way to drop the rows?
My data looks like:
item diff otherstuff
0 1 2 1
1 1 1 2
2 1 3 7
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
and should end up like:
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
but what I'm getting is:
item diff
0 1 1
1 2 -6
2 3 0
I've been looking through the documentation and can't find anything. I tried:
df1 = df.groupby(["item", "otherstuff"], as_index=false)["diff"].min()
df1 = df.groupby("item", as_index=false)["diff"].min()["otherstuff"]
df1 = df.groupby("item", as_index=false)["otherstuff", "diff"].min()
But none of those work (I realized with the last one that the syntax is meant for aggregating after a group is created).
Method #1: use idxmin() to get the indices of the elements of minimum diff, and then select those:
>>> df.loc[df.groupby("item")["diff"].idxmin()]
item diff otherstuff
1 1 1 2
6 2 -6 2
7 3 0 0
[3 rows x 3 columns]
Method #2: sort by diff, and then take the first element in each item group:
>>> df.sort_values("diff").groupby("item", as_index=False).first()
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
[3 rows x 3 columns]
Note that the resulting indices are different even though the row content is the same.
You can use DataFrame.sort_values with DataFrame.drop_duplicates:
df = df.sort_values(by='diff').drop_duplicates(subset='item')
print (df)
item diff otherstuff
6 2 -6 2
7 3 0 0
1 1 1 2
If possible multiple minimal values per groups and want all min rows use boolean indexing with transform for minimal values per groups:
print (df)
item diff otherstuff
0 1 2 1
1 1 1 2 <-multiple min
2 1 1 7 <-multiple min
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
print (df.groupby("item")["diff"].transform('min'))
0 1
1 1
2 1
3 -6
4 -6
5 -6
6 -6
7 0
8 0
Name: diff, dtype: int64
df = df[df.groupby("item")["diff"].transform('min') == df['diff']]
print (df)
item diff otherstuff
1 1 1 2
2 1 1 7
6 2 -6 2
7 3 0 0
The above answer worked great if there is / you want one min. In my case there could be multiple mins and I wanted all rows equal to min which .idxmin() doesn't give you. This worked
def filter_group(dfg, col):
return dfg[dfg[col] == dfg[col].min()]
df = pd.DataFrame({'g': ['a'] * 6 + ['b'] * 6, 'v1': (list(range(3)) + list(range(3))) * 2, 'v2': range(12)})
df.groupby('g',group_keys=False).apply(lambda x: filter_group(x,'v1'))
As an aside, .filter() is also relevant to this question but didn't work for me.
I tried everyone's method and I couldn't get it to work properly. Instead I did the process step-by-step and ended up with the correct result.
df.sort_values(by='item', inplace=True, ignore_index=True)
df.drop_duplicates(subset='diff', inplace=True, ignore_index=True)
df.sort_values(by=['diff'], inplace=True, ignore_index=True)
For a little more explanation:
Sort items by the minimum value you want
Drop the duplicates of the column you want to sort with
Resort the data because the data is still sorted by the minimum values
If you know that all of your "items" have more than one record you can sort, then use duplicated:
df.sort_values(by='diff').duplicated(subset='item', keep='first')
I'm using groupby on a pandas dataframe to drop all rows that don't have the minimum of a specific column. Something like this:
df1 = df.groupby("item", as_index=False)["diff"].min()
However, if I have more than those two columns, the other columns (e.g. otherstuff in my example) get dropped. Can I keep those columns using groupby, or am I going to have to find a different way to drop the rows?
My data looks like:
item diff otherstuff
0 1 2 1
1 1 1 2
2 1 3 7
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
and should end up like:
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
but what I'm getting is:
item diff
0 1 1
1 2 -6
2 3 0
I've been looking through the documentation and can't find anything. I tried:
df1 = df.groupby(["item", "otherstuff"], as_index=false)["diff"].min()
df1 = df.groupby("item", as_index=false)["diff"].min()["otherstuff"]
df1 = df.groupby("item", as_index=false)["otherstuff", "diff"].min()
But none of those work (I realized with the last one that the syntax is meant for aggregating after a group is created).
Method #1: use idxmin() to get the indices of the elements of minimum diff, and then select those:
>>> df.loc[df.groupby("item")["diff"].idxmin()]
item diff otherstuff
1 1 1 2
6 2 -6 2
7 3 0 0
[3 rows x 3 columns]
Method #2: sort by diff, and then take the first element in each item group:
>>> df.sort_values("diff").groupby("item", as_index=False).first()
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
[3 rows x 3 columns]
Note that the resulting indices are different even though the row content is the same.
You can use DataFrame.sort_values with DataFrame.drop_duplicates:
df = df.sort_values(by='diff').drop_duplicates(subset='item')
print (df)
item diff otherstuff
6 2 -6 2
7 3 0 0
1 1 1 2
If possible multiple minimal values per groups and want all min rows use boolean indexing with transform for minimal values per groups:
print (df)
item diff otherstuff
0 1 2 1
1 1 1 2 <-multiple min
2 1 1 7 <-multiple min
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
print (df.groupby("item")["diff"].transform('min'))
0 1
1 1
2 1
3 -6
4 -6
5 -6
6 -6
7 0
8 0
Name: diff, dtype: int64
df = df[df.groupby("item")["diff"].transform('min') == df['diff']]
print (df)
item diff otherstuff
1 1 1 2
2 1 1 7
6 2 -6 2
7 3 0 0
The above answer worked great if there is / you want one min. In my case there could be multiple mins and I wanted all rows equal to min which .idxmin() doesn't give you. This worked
def filter_group(dfg, col):
return dfg[dfg[col] == dfg[col].min()]
df = pd.DataFrame({'g': ['a'] * 6 + ['b'] * 6, 'v1': (list(range(3)) + list(range(3))) * 2, 'v2': range(12)})
df.groupby('g',group_keys=False).apply(lambda x: filter_group(x,'v1'))
As an aside, .filter() is also relevant to this question but didn't work for me.
I tried everyone's method and I couldn't get it to work properly. Instead I did the process step-by-step and ended up with the correct result.
df.sort_values(by='item', inplace=True, ignore_index=True)
df.drop_duplicates(subset='diff', inplace=True, ignore_index=True)
df.sort_values(by=['diff'], inplace=True, ignore_index=True)
For a little more explanation:
Sort items by the minimum value you want
Drop the duplicates of the column you want to sort with
Resort the data because the data is still sorted by the minimum values
If you know that all of your "items" have more than one record you can sort, then use duplicated:
df.sort_values(by='diff').duplicated(subset='item', keep='first')
I have a dataframe with points given in two columns x and y.
Thing x y length_x length_y
0 A 1 3 1 2
1 B 2 3 2 1
These (x,y) points are situated in the middle of one of the sides of a rectangle with vertex lengths length_x and length_y. What I wish to do is for each of these points give the coordinates of the rectangles they are on. That is: the following coordinated for Thing A would be:
(1+1*0.5, 3), (1-1*0.5,3), (1+1*0.5,3-2*0.5), (1-1*0.5, 3-2*0.5)
The half comes from the fact that the given lengths are the middle-points of an object so half the length is the distance from that point to the corner of the rectangle.
Hence my desired output is:
Thing x y Corner_x Corner_y length_x length_y
0 A 1 3 1.5 2.0 1 2
1 A 1 3 1.5 1.0 1 2
2 A 1 3 0.5 2.0 1 2
3 A 1 3 0.5 1.0 1 2
4 A 1 3 1.5 2.0 1 2
5 B 2 3 3.0 3.0 2 1
6 B 2 3 3.0 2.5 2 1
7 B 2 3 1.0 3.0 2 1
8 B 2 3 1.0 2.5 2 1
9 B 2 3 3.0 3.0 2 1
I tried to do this with defining a lambda returning two value but failed. Tried even to create multiple columns and then stack them, but it's really dirty.
bb = []
for thing in list_of_things:
new_df = df[df['Thing']=='{}'.format(thing)]
df = df.sort_values('x',ascending=False)
df['corner 1_x'] = df['x']+df['length_x']/2
df['corner 1_y'] = df['y']
df['corner 2_x'] = df['x']+1df['x_length']/2
df['corner 2_y'] = df['y']-df['length_y']/2
.........
Note also that the first corner's coordinates need to be repeated as I later what to use geopandas to transform each of these sets of coordinates into a POLYGON.
What I am looking for is a way to generate these rows is a fast and clean way.
You can use apply to create your corners as lists and explode them to the four rows per group.
Finally join the output to the original dataframe:
df.join(df.apply(lambda r: pd.Series({'corner_x': [r['x']+r['length_x']/2, r['x']-r['length_x']/2],
'corner_y': [r['y']+r['length_y']/2, r['y']-r['length_y']/2],
}), axis=1).explode('corner_x').explode('corner_y'),
how='right')
output:
Thing x y length_x length_y corner_x corner_y
0 A 1 3 1 2 1.5 4
0 A 1 3 1 2 1.5 2
0 A 1 3 1 2 0.5 4
0 A 1 3 1 2 0.5 2
1 B 2 3 2 1 3 3.5
1 B 2 3 2 1 3 2.5
1 B 2 3 2 1 1 3.5
1 B 2 3 2 1 1 2.5
The problem consist on calculate from a dataframe the column "accumulated" using the columns "accumulated" and "weekly". The formula to do this is accumulated in t = weekly in t + accumulated in t-1
The desired result should be:
weekly accumulated
2 0
1 1
4 5
2 7
The result I'm obtaining is:
weekly accumulated
2 0
1 1
4 4
2 2
What I have tried is:
for key, value in df_dic.items():
df_aux = df_dic[key]
df_aux['accumulated'] = 0
df_aux['accumulated'] = (df_aux.weekly + df_aux.accumulated.shift(1))
#df_aux["accumulated"] = df_aux.iloc[:,2] + df_aux.iloc[:,3].shift(1)
df_aux.iloc[0,3] = 0 #I put this because I want to force the first cell to be 0.
Being df_aux.iloc[0,3] the first row of the column "accumulated".
What I´m doing wrong?
Thank you
EDIT: df_dic is a dictionary with 5 dataframes. df_dic is seen as {0: df1, 1:df2, 2:df3}. All the dataframes have the same size and same columns names. So i do the for loop to do the same calculation in every dataframe inside the dictionary.
EDIT2 : I'm trying doing the computation outside the for loop and is not working.
What im doing is:
df_auxp = df_dic[0]
df_auxp['accumulated'] = 0
df_auxp['accumulated'] = df_auxp["weekly"] + df_auxp["accumulated"].shift(1)
df_auxp.iloc[0,3] = df_auxp.iloc[0,3].fillna(0)
Maybe have something to do with the dictionary interaction...
To solve for 3 dataframes
import pandas as pd
df1 = pd.DataFrame({'weekly':[2,1,4,2]})
df2 = pd.DataFrame({'weekly':[3,2,5,3]})
df3 = pd.DataFrame({'weekly':[4,3,6,4]})
print (df1)
print (df2)
print (df3)
for d in [df1,df2,df3]:
d['accumulated'] = d['weekly'].cumsum() - d.iloc[0,0]
print (d)
The output of this will be as follows:
Original dataframes:
df1
weekly
0 2
1 1
2 4
3 2
df2
weekly
0 3
1 2
2 5
3 3
df3
weekly
0 4
1 3
2 6
3 4
Updated dataframes:
df1:
weekly accumulated
0 2 0
1 1 1
2 4 5
3 2 7
df2:
weekly accumulated
0 3 0
1 2 2
2 5 7
3 3 10
df3:
weekly accumulated
0 4 0
1 3 3
2 6 9
3 4 13
To solve for 1 dataframe
You need to use cumsum and then subtract the value from first row. That will give you the desired result. here's how to do it.
import pandas as pd
df = pd.DataFrame({'weekly':[2,1,4,2]})
print (df)
df['accumulated'] = df['weekly'].cumsum() - df.iloc[0,0]
print (df)
Original dataframe:
weekly
0 2
1 1
2 4
3 2
Updated dataframe:
weekly accumulated
0 2 0
1 1 1
2 4 5
3 2 7
import pandas as pd
x = pd.DataFrame([[1,2,3],[4,5,6]])
x[::2]
what does the above command mean and how does it function?
Better is more data, it return even rows only by slicing:
x = pd.DataFrame([[1,2,3],[4,5,6],[7,8,9],[0,1,2]])
print (x)
0 1 2
0 1 2 3
1 4 5 6
2 7 8 9
3 0 1 2
print (x[::2])
0 1 2
0 1 2 3
2 7 8 9