I have a dataframe and would like to assign multiple values from one row to multiple other rows.
I get it to work with .iloc but for some when I use conditions with .loc it only returns nan.
df = pd.DataFrame(dict(A = [1,2,0,0],B=[0,0,0,10],C=[3,4,5,6]))
df.index = ['a','b','c','d']
When I use loc with conditions or with direct index names:
df.loc[df['A']>0, ['B','C']] = df.loc['d',['B','C']]
df.loc[['a','b'], ['B','C']] = df.loc['d',['B','C']]
it will return
A B C
a 1.0 NaN NaN
b 2.0 NaN NaN
c 0.0 0.0 5.0
d 0.0 10.0 6.0
but when I use .iloc it actually works as expected
df.iloc[0:2,1:3] = df.iloc[3,1:3]
A B C
a 1 10 6
b 2 10 6
c 0 0 5
d 0 10 6
is there a way to do this with .loc or do I need to rewrite my code to get the row numbers from my mask?
When you use labels, pandas perform index alignment, and in your case there is no common indices thus the NaNs, while location based indexing does not align.
You can assign a numpy array to prevent index alignment:
df.loc[['a','b'], ['B','C']] = df.loc['d',['B','C']].values
output:
A B C
a 1 10 6
b 2 10 6
c 0 0 5
d 0 10 6
Related
I have the following pandas dataframe and would like to build a new column 'c' which is the summation of column 'b' value and column 'a' previous values. With shifting column 'a' it is possible to do so. However, I would like to know how I can pass the previous values of column 'a' in the apply() function.
l1 = [1,2,3,4,5]
l2 = [3,2,5,4,6]
df = pd.DataFrame(data=l1, columns=['a'])
df['b'] = l2
df['shifted'] = df['a'].shift(1)
df['c'] = df.apply(lambda row: row['shifted']+ row['b'], axis=1)
print(df)
a b shifted c
0 1 3 NaN NaN
1 2 2 1.0 3.0
2 3 5 2.0 7.0
3 4 4 3.0 7.0
4 5 6 4.0 10.0
I appreciate your help.
Edit: this is a dummy example. I need to use the apply function because I'm passing another function to it which uses previous rows of some columns and checks some condition.
First let's make it clear that you do not need apply for this simple operation, so I'll consider it as a dummy example of a complex function.
Assuming non-duplicate indices, you can generate a shifted Series and reference it in apply using the name attribute:
s = df['a'].shift(1)
df['c'] =df.apply(lambda row: row['b']+s[row.name], axis=1)
output:
a b shifted c
0 1 3 NaN NaN
1 2 2 1.0 3.0
2 3 5 2.0 7.0
3 4 4 3.0 7.0
4 5 6 4.0 10.0
I have a smiliar question to this one.
I have a dataframe with several rows, which looks like this:
Name TypA TypB ... TypF TypA_value TypB_value ... TypF_value Divider
1 1 1 NaN 10 5 NaN 5
2 NaN 2 NaN NaN 20 NaN 10
and I want to divide all columns with the ending "value" by the column "Divider", how can I do so? One trick would be to use the sorting, to use the answer from above, but is there a direct way for it? That I do not need to sort the dataframe.
The outcome would be:
Name TypA TypB ... TypF TypA_value TypB_value ... TypF_value Divider
1 1 1 NaN 2 1 0 5
2 NaN 2 NaN 0 2 0 10
So a NaN will lead to a 0.
Use DataFrame.filter to filter the columns like value from dataframe then use DataFrame.div along axis=0 to divide it by column Divider, finally use DataFrame.update to update the values in dataframe:
d = df.filter(like='_value').div(df['Divider'], axis=0).fillna(0)
df.update(d)
Result:
Name TypA TypB TypF TypA_value TypB_value TypF_value Divider
0 1 1.0 1 NaN 2.0 1.0 0.0 5
1 2 NaN 2 NaN 0.0 2.0 0.0 10
You could select the columns of interest using DataFrame.filter, and divide as:
value_cols = df.filter(regex=r'_value$').columns
df[value_cols] /= df['Divider'].to_numpy()[:,None]
# df[value_cols] = df[value_cols].fillna(0)
print(df)
Name TypA TypB TypF TypA_value TypB_value TypF_value Divider
0 1 1.0 1 NaN 2.0 1.0 NaN 5
1 2 NaN 2 NaN NaN 2.0 NaN 10
Taking two sample columns A and B :
import pandas as pd
import numpy as np
a={ 'Name':[1,2],
'TypA':[1,np.nan],
'TypB':[1,2],
'TypA_value':[10,np.nan],
'TypB_value':[5,20],
'Divider':[5,10]
}
df = pd.DataFrame(a)
cols_all = df.columns
Find columns for which calculations are to be done. Assuming there all have 'value' and an underscore :
cols_to_calc = [c for c in cols_all if '_value' in c]
For these columns: first, divide with the divider column then replace nan with 0 in those columns.
for c in cols_to_calc:
df[c] = df[c] / df.Divider
df[c] = df[c].fillna(0)
I have a df:
a b c
1 2 3 6
2 2 5 7
3 4 6 8
I want every nth row of groupby a:
w=df.groupby('a').nth(0) #first row
x=df.groupby('a').nth(1) #second row
The second group of the df has no second row, in this case I want to have 'None' values.
[In:] df.groupby('a').nth(1)
[Out:]
a b c
1 2 5 7
2 None None None
Or maybe simplier:
The df has 1-4 rows within groups. If a group has less than 4 rows, I want to extend the group, so that it has 4 rows and fill the missing rows with 'None'. Afterwards if I pick the nth row of groups, I have the desired output.
If you are just interested in a specific nth but not have enough rows in some groups, you can consider to use reindex with unique value from the column a like:
print (df.groupby('a').nth(1).reindex(df['a'].unique()).reset_index())
a b c
0 2 5.0 7.0
1 4 NaN NaN
One way is to assign a count/rank column and reindex/stack:
n=2
(df.assign(rank=df.groupby('a').cumcount())
.query(f'rank < #n')
.set_index(['a','rank'])
.unstack('rank')
.stack('rank', dropna=False)
.reset_index()
.drop('rank', axis=1)
)
Output:
a b c
0 2 3.0 6.0
1 2 5.0 7.0
2 4 6.0 8.0
3 4 NaN NaN
I have a df with values:
A B C D
0 1 2 3 2
1 2 3 3 9
2 5 3 6 6
3 3 6 7
4 6 7
5 2
df.shape is 6x4, say
df.iloc[:,1] pulls out the B column, but len(df.iloc[:,1]) is also = 6
How do I "reshape" df.iloc[:,1]? Which function can I use so that the output is the length of the actual values in the column.
My expected output in this case is 3
You can use last_valid_index. Just note that since your series originally contained NaN values and these are considered float, even after filtering your series will be float. You may wish to convert to int as a separate step.
# first convert dataframe to numeric
df = df.apply(pd.to_numeric, errors='coerce')
# extract column
B = df.iloc[:, 1]
# filter to the last valid value
B_filtered = B[:B.last_valid_index()]
print(B_filtered)
0 2.0
1 3.0
2 3.0
3 6.0
Name: B, dtype: float64
You can use list comprehension like this.
len([x for x in df.iloc[:,1] if x != ''])
I am trying to do the following: on a dataframe X, I want to select all rows where X['a']>0 but I want to preserve the dimension of X, so that any other row will appear as containing NaN. Is there a fast way to do it? If one does X[X['a']>0] the dimensions of X are not preserved.
Use double subscript [[]]:
In [42]:
df = pd.DataFrame({'a':np.random.randn(10)})
df
Out[42]:
a
0 1.042971
1 0.978914
2 0.764374
3 -0.338405
4 0.974011
5 -0.995945
6 -1.649612
7 0.965838
8 -0.142608
9 -0.804508
In [48]:
df[df[['a']] > 1]
Out[48]:
a
0 1.042971
1 NaN
2 NaN
3 NaN
4 NaN
5 NaN
6 NaN
7 NaN
8 NaN
9 NaN
The key semantic difference here is what is returned is a df when you double subscript so this masks the df itself rather than the index
Note though that if you have multiple columns then it will mask all those as NaN