I have a dataframe as follow:
import pandas as pd
d = {'location1': [1, 2,3,8,6], 'location2':
[2,1,4,6,8]}
df = pd.DataFrame(data=d)
The dataframe df means there is a road between two locations. look like:
location1 location2
0 1 2
1 2 1
2 3 4
3 8 6
4 6 8
The first row means there is a road between locationID1 and locationID2, however, the second row also encodes this information. The forth and fifth rows also have repeated information. I am trying the remove those repeated by keeping only one row. Any of row is okay.
For example, my expected output is
location1 location2
0 1 2
2 3 4
4 6 8
Any efficient way to do that because I have a large dataframe with lots of repeated rows.
Thanks a lot,
It looks like you want every other row in your dataframe. This should work.
import pandas as pd
d = {'location1': [1, 2,3,8,6], 'location2':
[2,1,4,6,8]}
df = pd.DataFrame(data=d)
print(df)
location1 location2
0 1 2
1 2 1
2 3 4
3 8 6
4 6 8
def Every_other_row(a):
return a[::2]
Every_other_row(df)
location1 location2
0 1 2
2 3 4
4 6 8
Related
I have two dataframes and one is a subset of another one (picture below). I am not sure whether pandas can compare two dataframes and filter the data which is not in the subset and export it as a dataframe. Or is there any package doing this kind of task?
The subset dataframe was generated from RandomUnderSampler but the RandomUnderSampler did not have function which exports the unselected data. Any comments are welcome.
Use drop_duplicates with keep=False parameter:
Example:
>>> df1
A B
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9
>>> df2
A B
0 0 1
1 2 3
2 6 7
>>> pd.concat([df1, df2]).drop_duplicates(keep=False)
A B
2 4 5
4 8 9
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
I have a pandas dataframe as below:
import pandas as pd
df = pd.DataFrame({'ORDER':["A", "A", "A", "B", "B","B"], 'GROUP': ["A_2018_1B1", "A_2018_1B1", "A_2018_1M1", "B_2018_I000_1C1", "B_2018_I000_1B1", "B_2018_I000_1C1H"], 'VAL':[1,3,8,5,8,10]})
df
ORDER GROUP VAL
0 A A_2018_1B1 1
1 A A_2018_1B1H 3
2 A A_2018_1M1 8
3 B B_2018_I000_1C1 5
4 B B_2018_I000_1B1 8
5 B B_2018_I000_1C1H 10
I want to create a column "CAL" as sum of 'VAL' where GROUP name is same for all the rows expect H character in the end. So, for example, 'VAL' column for 1st two rows will be added because the only difference between the 'GROUP' is 2nd row has H in the last. Row 3 will remain as it is, Row 4 and 6 will get added and Row 5 will remain same.
My expected output
ORDER GROUP VAL CAL
0 A A_2018_1B1 1 4
1 A A_2018_1B1H 3 4
2 A A_2018_1M1 8 8
3 B B_2018_I000_1C1 5 15
4 B B_2018_I000_1B1 8 8
5 B B_2018_I000_1C1H 10 15
Try with replace then transform
df.groupby(df.GROUP.str.replace('H','')).VAL.transform('sum')
0 4
1 4
2 8
3 15
4 8
5 15
Name: VAL, dtype: int64
df['CAL'] = df.groupby(df.GROUP.str.replace('H','')).VAL.transform('sum')
Suppose I have dataframe like this
>>> df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4],'value':[1,2,3,1,2,3,4,1,1]})
>>> df
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
Now I want top all records from each group using group id except last 3. That means I want to drop last 3 records from all groups. How can I do it using pandas group_by. This is dummy data.
Use GroupBy.cumcount for counter from back by ascending=False and then compare by Series.gt for greater values like 2, because python count from 0:
df = df[df.groupby('id').cumcount(ascending=False).gt(2)]
print (df)
id value
3 2 1
Details:
print (df.groupby('id').cumcount(ascending=False))
0 2
1 1
2 0
3 3
4 2
5 1
6 0
7 0
8 0
dtype: int64
I have this datafram
PNN_sh NN_shap PNN_corr NN_corr
1 25005 1 25005
2 25012 2 25001
3 25011 3 25009
4 25397 4 25445
5 25006 5 25205
Then I made 2 dataframs from this one.
NN_sh = data[['PNN_sh', 'NN_shap']]
NN_corr = data[['PNN_corr', 'NN_corr']]
Thereafter, I sorted them and saved in new dataframes.
NN_sh_sort = NN_sh.sort_values(by=['NN_shap'])
NN_corr_sort = NN_corr.sort_values(by=['NN_corr'])
Now I want to combine 2 columns from the 2 dataframs above.
all_pd = pd.concat([NN_sh_sort['PNN_sh'], NN_corr_sort['PNN_corr']], axis=1, join='inner')
But what I got is only the first column copied into second one also.
PNN_sh PNN_corr
1 1
5 5
3 3
2 2
4 4
The second column should be
PNN_corr
2
1
3
5
4
Any idea how to fix it? Thanks in advance
Put ignore_index=True to sort_values():
NN_sh_sort = NN_sh.sort_values(by=['NN_shap'], ignore_index=True)
NN_corr_sort = NN_corr.sort_values(by=['NN_corr'], ignore_index=True)
Then the result after concat will be:
PNN_sh PNN_corr
0 1 2
1 5 1
2 3 3
3 2 5
4 4 4
I think when you sort you are preserving the original indices of the example DataFrames. Therefore, it is joining the PNN_corr value that was originally in the same row (at same index). Try resetting the index of each DataFrame after sorting, then join/concat.
NN_sh_sort = NN_sh.sort_values(by=['NN_shap']).reset_index()
NN_corr_sort = NN_corr.sort_values(by=['NN_corr']).reset_index()
all_pd = pd.concat([NN_sh_sort['PNN_sh'], NN_corr_sort['PNN_corr']], axis=1, join='inner')