I have a dataframe like below in which I need to replace the 0s with the mean of the rows where the parent_key matches the self_key.
Input DataFrame: df= pd.DataFrame ({'self_key':['a','b','c','d','e','e','e','f','f','f'],'parent_key':[np.nan,'a','b','b','c','c','c','d','d','d'], 'value':[0,0,0,0,4,6,14,12,8,22],'level':[1,2,3,3,4,4,4,4,4,4]})
The row 3 has self_key of 'd' so I would need to replace its 0 value in column 'value' with the mean of rows 7,8,9 to fill with the correct value of 14. Since the lower levels feed into the higher levels I would need to do it from lowest level to highest to fill out the dataframe as well but when I do the below code it doesn't work and I get the error "ValueError: Grouper for '<class 'pandas.core.frame.DataFrame'>' not 1-dimensional". How can I fill in the 0s with the means from lowest level to highest?
df['value']=np.where((df['value']==0) & (df['level']==3), df['value'].groupby(df.where(df['parent_key']==df['self_key'])).transform('mean'), df['value'])
Input
self_key parent_key value level
0 a NaN 0 1
1 b a 0 2
2 c b 0 3
3 d b 0 3
4 e c 4 4
5 e c 6 4
6 e c 14 4
7 f d 12 4
8 f d 8 4
9 f d 22 4
My approach is to repeat the above code 3 times and change the level from 3 to 2 to 1, but its not working for even level 3.
Expected Ouput:
self_key parent_key value level
0 a NaN 11 1
1 b a 11 2
2 c b 8 3
3 d b 14 3
4 e c 4 4
5 e c 6 4
6 e c 14 4
7 f d 12 4
8 f d 8 4
9 f d 22 4
If I understand your problem correctly, you are trying to compute mean in a bottom-up fashion by filtering dataframe on certain keys. If so, then following should solve it:
for l in range(df["level"].max()-1, 0, -1):
df_sub = df[(df["level"] == l) & (df["value"] == 0)]
self_keys = df_sub["self_key"].tolist()
for k in self_keys:
df.loc[df_sub[df_sub["self_key"] == k].index, "value"] = df[df["parent_key"] == k]["value"].mean()
[Out]:
self_key parent_key value level
0 a 11 1
1 b a 11 2
2 c b 8 3
3 d b 14 3
4 e c 4 4
5 e c 6 4
6 e c 14 4
7 f d 12 4
8 f d 8 4
9 f d 22 4
Related
I am having issues trying to modify a value if columnA contains lets say 0,2,12,44,75,81 (looking at 33 different numerical values in columnA. If columnA contains 1 of the 33 vaules defined, i need to then change the same row of colulmnB to "approved".
I have tried this:
df[(df.columnA == '0|2|12|44|75|81')][df.columnB] = 'approved'
I get an error that there are none of index are in the columns but, i know that isn't correct looking at my csv file data. I must be searching wrong or using the wrong syntax.
As others have said, if you need the value in column A to match exactly with a value in the list, then you can use .isin. If you need a partial match, then you can convert to string dtypes and use .contains.
setup:
nums_to_match = [0,2,9]
df
A B
0 6 e
1 1 f
2 3 b
3 6 g
4 6 i
5 0 f
6 9 a
7 6 i
8 6 a
9 2 h
Exact match:
df.loc[df.A.isin(nums_to_match),'B']='approved'
df:
A B
0 6 e
1 1 f
2 3 b
3 6 g
4 6 i
5 0 approved
6 9 approved
7 6 i
8 6 a
9 2 approved
partial match:
nums_to_match_str = list(map(str,nums_to_match))
df['A']=df['A'].astype(str)
df.loc[df.A.str.contains('|'.join(nums_to_match_str),case=False,na=False),'B']='approved'
df
A B
0 1 h
1 4 c
2 6 i
3 7 d
4 3 d
5 9 approved
6 5 i
7 1 c
8 0 approved
9 5 d
Given two data frames. One contains a column of repeated values (a, in this case). The other contains what this value corresponds to (in this example, it corresponds to some "d" values). How do I efficiently replenish the first data frame with a new column, values in which correspond to some existent column, according to a rule recorded in the other data frame. Here is an example code that works really slow:
import pandas as pd
import numpy as np
d1 = pd.DataFrame(np.asarray([[1,2,3], [2,4,5], [3,4,5], [2,1,4], [3,4,5]]), columns = ['a', 'b', 'c'])
d2 = pd.DataFrame(np.asarray([[1,7], [2,8], [3,11]]), columns = ['a', 'd'])
d = np.empty((d1.shape[0],))
for i in range(d1.shape[0]):
temp = d2.loc[d2['a'] == d1.at[i,'a']]
d[i] = temp['d'].array[0]
d1['d'] = d
This is d1 original:
a b c
0 1 2 3
1 2 4 5
2 3 4 5
3 2 1 4
4 3 4 5
This is d2:
a d
0 1 7
1 2 8
2 3 11
This is a resultant d1:
a b c d
0 1 2 3 7
1 2 4 5 8
2 3 4 5 11
3 2 1 4 8
4 3 4 5 11
You're probably looking for pd.merge.
In your case, d1 = d1.merge(d2, on=['a'], how='left') should do the trick.
Another way is to use map and make only the values you need.
d1['d'] = d1['a'].map(d2.set_index('a')['d'])
d1
Output:
a b c d
0 1 2 3 7
1 2 4 5 8
2 3 4 5 11
3 2 1 4 8
4 3 4 5 11
i have below df
B C D E
2 2 4 11
11 0 5 3
12 10 1 11
5 9 7 15
1st i wants a unique value from whole df like below:
[0,1,2,3,4,5,7,9,10,11,12,15]
then i wants final output
value value exists in number of col
0 1
1 1
2 2
3 1
4 1
5 1
7 1
9 1
10 1
11 2
12 1
15 1
that means each value,how many columns its available
i wants that output
Using python you can do something like this:
# your input df as a list of lists
df = [[2,11,12,5], [2,0,10,9], [4,5,1,7], [11,3,11,15]]
#remove duplicates in each list
dfU = [list(set(l)) for l in df]
# sort each list (not required for this approach)
for l in dfU:
l.sort()
# the requested unique list
flatList = [item for sublist in df for item in sublist]
uniqueList = list(set(flatList))
print(uniqueList)
# output as a list of lists
output = []
for num in uniqueList:
cnt = 0
for idx in range(len(dfU)):
if dfU[idx].count(num) > 0:
cnt+=1
output.append([num,cnt])
print(output)
Side note, the count function is computationally expensive, so it would be better to do a linear scan along all sorted columns.
Use DataFrame.melt for reshape, remove duplicates by both columns and count by GroupBy.size with Series.reset_index for DataFrame:
df1 = (df.melt(value_name='value')
.drop_duplicates()
.groupby('value')
.size()
.reset_index(name='count'))
print (df1)
value count
0 0 1
1 1 1
2 2 2
3 3 1
4 4 1
5 5 2
6 7 1
7 9 1
8 10 1
9 11 2
10 12 1
11 15 1
Details:
print (df.melt(value_name='value'))
variable value
0 B 2
1 B 11
2 B 12
3 B 5
4 C 2
5 C 0
6 C 10
7 C 9
8 D 4
9 D 5
10 D 1
11 D 7
12 E 11
13 E 3
14 E 11
15 E 15
One 11 for index 14 is removed:
print (df.melt(value_name='value').drop_duplicates())
variable value
0 B 2
1 B 11
2 B 12
3 B 5
4 C 2
5 C 0
6 C 10
7 C 9
8 D 4
9 D 5
10 D 1
11 D 7
12 E 11
13 E 3
15 E 15
If want pure python solution:
from collections import Counter
L = sorted(Counter([y for x in df.T.values for y in set(x)]).items())
df1 = pd.DataFrame(L, columns=['value','count'])
print (df1)
value count
0 0 1
1 1 1
2 2 2
3 3 1
4 4 1
5 5 2
6 7 1
7 9 1
8 10 1
9 11 2
10 12 1
11 15 1
I have this data
ID Value1 Value2 Type Type2
1 3 1 A X
2 2 2 A X
3 5 3 B Y
4 2 4 B Z
5 6 8 C Z
6 7 9 C Z
7 8 0 C L
8 3 2 D M
9 4 3 D M
10 6 5 D M
11 8 7 D M
Right now i am able to generate this output using this code
pandabook.groupby(['Type','Type2'],as_index=False)['Value1', 'Value2'].agg({'Value1': 'sum','Value2': 'sum'})
ID Value 1 Value2 Type Type2
1 5 3 A X
2 5 3 B Y
3 2 5 B Z
4 13 17 C Z
5 8 0 C L
6 21 17 D M
I want to show the Aggregated count as well, as show in this example
How can i achieve this output ?
Add new value to dictionary with size function, remove as_index=False for prevent:
ValueError: cannot insert Type, already exists
and last rename with reset_index:
df = pandabook.groupby(['Type','Type2']).agg({'Value1': 'sum','Value2': 'sum', 'Type':'size'})
df = df.rename(columns={'Type':'Count'}).reset_index()
print (df)
Type Type2 Value1 Value2 Count
0 A X 5 3 2
1 B Y 5 3 1
2 B Z 2 4 1
3 C L 8 0 1
4 C Z 13 17 2
5 D M 21 17 4
I currently have a pandas dataframe where values between 0 and 1 are saved. I am looking for a function which can provide me the top 5 values of a column, together with the name of the column and the associated index of the values.
Sample Input: data frame with column names a:z, index 1:23, entries are values between 0 and 1
Sample Output: array of 5 highest entries in each column, each with column name and index
Edit:
For the following data frame:
np.random.seed([3,1415])
df = pd.DataFrame(np.random.randint(10, size=(10, 4)), list('abcdefghij'), list('ABCD'))
df
A B C D
a 0 2 7 3
b 8 7 0 6
c 8 6 0 2
d 0 4 9 7
e 3 2 4 3
f 3 6 7 7
g 4 5 3 7
h 5 9 8 7
i 6 4 7 6
j 2 6 6 5
I would like to get an output like (for example for the first column):
[[8,b,A], [8, c, A], [6,i,A], [5, h, A], [4,g,A]].
consider the dataframe df
np.random.seed([3,1415])
df = pd.DataFrame(
np.random.randint(10, size=(10, 4)), list('abcdefghij'), list('ABCD'))
df
A B C D
a 0 2 7 3
b 8 7 0 6
c 8 6 0 2
d 0 4 9 7
e 3 2 4 3
f 3 6 7 7
g 4 5 3 7
h 5 9 8 7
i 6 4 7 6
j 2 6 6 5
I'm going to use np.argpartition to separate each column into the 5 smallest and 10 - 5 (also 5) largest
v = df.values
i = df.index.values
k = len(v) - 5
pd.DataFrame(
i[v.argpartition(k, 0)[-k:]],
np.arange(k), df.columns
)
A B C D
0 g f i i
1 b c a d
2 h h f h
3 i b d f
4 c j h g
print(your_dataframe.sort_values(ascending=False)[0:4])