I want to find the argmax of the values in a matrix by column, e.g.:
1 2 3 2 3 3
4 5 6 ->
3 7 8
I feel like I should just be able to map an argmax/posmax function over the columns, but I don't see a particularly intuitive way to do this in Octave.
Read max function documentation here
[max_values indices] = max(input);
Example:
input =
1 2 3
4 5 6
3 7 8
[max_values indices] = max(input)
max_values =
4 7 8
indices =
2 3 3
In Octave If
A =
1 3 2
6 5 4
7 9 8
1) For Each Column Max value and corresponding index of them can be found by
>> [max_values,indices] =max(A,[],1)
max_values =
7 9 8
indices =
3 3 3
2) For Each Row Max value and corresponding index of them can be found by
>> [max_values,indices] =max(A,[],2)
max_values =
3
6
9
indices =
2
1
2
Similarly For minimum value
>> [min_values,indices] =min(A,[],1)
min_values =
1 3 2
indices =
1 1 1
>> [min_values,indices] =min(A,[],2)
min_values =
1
4
7
indices =
1
3
1
Related
Starting Dataframe:
A B
0 1 1
1 1 2
2 2 3
3 3 4
4 3 5
5 1 6
6 1 7
7 1 8
8 2 9
Desired result - eg. Remove rows where column A has values that match the row above or below:
A B
0 1 1
2 2 3
3 3 4
5 1 6
8 2 9
You can use boolean indexing, the following condition will return true if value of A is NOT equal to value of A's next row
new_df = df[df['A'].ne(df['A'].shift())]
A B
0 1 1
2 2 3
3 3 4
5 1 6
8 2 9
I have a dataframe
Intialise data of lists.
data = {'Id':['1', '2', '3', '4','5','6','7','8','9','10'], 'reply_id':[2, 2,2, 5,5,6,8,8,1,1]}
Create DataFrame
df = pd.DataFrame(data)
Id reply_id
0 1 2
1 2 2
2 3 2
3 4 5
4 5 5
5 6 6
6 7 8
7 8 8
8 9 1
9 10 1
I want to get total of reply_id in new for every Id.
Id=1 have 2 time occurrence in reply_id which i want in new column new
Desired output
Id reply_id new
0 1 2 2
1 2 2 3
2 3 2 0
3 4 5 0
4 5 5 2
5 6 6 1
6 7 8 0
7 8 8 2
8 9 1 0
9 10 1 0
I have done this line of code.
df['new'] = df.reply_id.eq(df.Id).astype(int).groupby(df.Id).transform('sum')
In this answer, I used Series.value_counts to count values in reply_id, and converted the result to a dict. Then, I used Series.map on the Id column to associate counts to Id. fillna(0) is used to fill values not present in reply_id
df['new'] = (df['Id']
.astype(int)
.map(df['reply_id'].value_counts().to_dict())
.fillna(0)
.astype(int))
Use, Series.groupby on the column reply_id, then use the aggregation function GroupBy.count to create a mapping series counts, finally use Series.map to map the values in Id column with their respective counts:
counts = df['reply_id'].groupby(df['reply_id']).count()
df['new'] = df['Id'].map(counts).fillna(0).astype(int)
Result:
# print(df)
Id reply_id new
0 1 2 2
1 2 2 3
2 3 2 0
3 4 5 0
4 5 5 2
5 6 6 1
6 7 8 0
7 8 8 2
8 9 1 0
9 10 1 0
I have a data set like this:-
S.No.,Year of birth,year of death
1, 1, 5
2, 3, 6
3, 2, -
4, 5, 7
I need to calculate population on till that years let say:-
year,population
1 1
2 2
3 3
4 3
5 4
6 3
7 2
8 1
How can i solve it in pandas?
Since i am not good in pandas.
Any help would be appreciate.
First is necessary choose maximum year of year of death if not exist, in solution is used 8.
Then convert values of year of death to numeric and replace missing values by this year. In first solution is used difference between birth and death column with Index.repeat with GroupBy.cumcount, for count is used Series.value_counts:
#if need working with years
#today_year = pd.to_datetime('now').year
today_year = 8
df['year of death'] = pd.to_numeric(df['year of death'], errors='coerce').fillna(today_year)
df = df.loc[df.index.repeat(df['year of death'].add(1).sub(df['Year of birth']).astype(int))]
df['Year of birth'] += df.groupby(level=0).cumcount()
df1 = (df['Year of birth'].value_counts()
.sort_index()
.rename_axis('year')
.reset_index(name='population'))
print (df1)
year population
0 1 1
1 2 2
2 3 3
3 4 3
4 5 4
5 6 3
6 7 2
7 8 1
Another solution use list comprehension with range for repeat years:
#if need working with years
#today_year = pd.to_datetime('now').year
today_year = 8
s = pd.to_numeric(df['year of death'], errors='coerce').fillna(today_year)
L = [x for s, e in zip(df['Year of birth'], s) for x in range(s, e + 1)]
df1 = (pd.Series(L).value_counts()
.sort_index()
.rename_axis('year')
.reset_index(name='population'))
print (df1)
year population
0 1 1
1 2 2
2 3 3
3 4 3
4 5 4
5 6 3
6 7 2
7 8 1
Similar like before, only is used Counter for dictionary for final DataFrame:
from collections import Counter
#if need working with years
#today_year = pd.to_datetime('now').year
today_year = 8
s = pd.to_numeric(df['year of death'], errors='coerce').fillna(today_year)
d = Counter([x for s, e in zip(df['Year of birth'], s) for x in range(s, e + 1)])
print (d)
Counter({5: 4, 3: 3, 4: 3, 6: 3, 2: 2, 7: 2, 1: 1, 8: 1})
df1 = pd.DataFrame({'year':list(d.keys()),
'population':list(d.values())})
print (df1)
year population
0 1 1
1 2 2
2 3 3
3 4 3
4 5 4
5 6 3
6 7 2
7 8 1
I need to create a incremental series for a given value of dataframe in python.
Any help much appreciated
Suppose I have dataframe column
df['quadrant']
Out[6]:
0 4
1 4
2 4
3 3
4 3
5 3
6 2
7 2
8 2
9 1
10 1
11 1
I want to create a new column such that
index quadrant new value
0 4 1
1 4 5
2 4 9
3 3 2
4 3 6
5 3 10
6 2 3
7 2 7
8 2 11
9 1 4
10 1 8
11 1 12
Using Numpy, you can create the array as:
import numpy as np
def value(q, k=1):
diff_quadrant = np.diff(q)
j = 0
ramp = []
for i in np.where(diff_quadrant != 0)[0]:
ramp.extend(list(range(i-j+1)))
j = i+1
ramp.extend(list(range(len(quadrant)-j)))
ramp = np.array(ramp) * k # sawtooth-shaped array
a = np.ones([len(quadrant)], dtype = np.int)*5
return a - q + ramp
quadrant = np.array([3, 3, 3, 3, 4, 4, 4, 2, 2, 1, 1, 1])
b = value(quadrant, 4)
# [ 2 6 10 14 1 5 9 3 7 4 8 12]
I am going to convert an array as [1 2 6 12 13 15 20 8 30 31] to range [1 2 3 4 5 6 7 8 9 10] in EXCEL or in MATLAB. Is there any function to do it? Is there any suggestion to convert this type of arbitrary range to definite range?
Solution for MATLAB
In MATLAB, unique can do the trick. For example:
A = [1 1 1 1 4 4 5 3 3 6 6 6 6 8]
[C, ia, ic] = unique(A);
Now ic holds the values that you're looking for:
ic =
1 1 1 1 3 3 4 2 2 5 5 5 5 6
Also, you can remap ic to a new order if you wish to do so. For instance, here's a new order similar to the one you suggested:
new_order = [1 3 2 4 6 5];
ic_new = new_order(ic);
which yields:
ic_new =
1 1 1 1 2 2 4 3 3 6 6 6 6 5
Hope that helps!
In Matlab:
v = [1 2 6 12 13 15 20 8 30 31];
v_index = 1:length(v);