Assume, there are two pandas DataFrame: df1 & df2. The df1 is a square data frame such as following
import numpy as np
import pandas as pd
item_names = [2,7,9,10,11,13,14,21,24]
np.random.seed(123)
nums = np.round(np.random.random(size=(9,9)),2)
df1 = pd.DataFrame(nums, index=item_names, columns=item_names)
df1 output:
2 7 9 10 11 13 14 21 24
2 0.70 0.29 0.23 0.55 0.72 0.42 0.98 0.68 0.48
7 0.39 0.34 0.73 0.44 0.06 0.40 0.74 0.18 0.18
9 0.53 0.53 0.63 0.85 0.72 0.61 0.72 0.32 0.36
10 0.23 0.29 0.63 0.09 0.43 0.43 0.49 0.43 0.31
11 0.43 0.89 0.94 0.50 0.62 0.12 0.32 0.41 0.87
13 0.25 0.48 0.99 0.52 0.61 0.12 0.83 0.60 0.55
14 0.34 0.30 0.42 0.68 0.88 0.51 0.67 0.59 0.62
21 0.67 0.84 0.08 0.76 0.24 0.19 0.57 0.10 0.89
24 0.63 0.72 0.02 0.59 0.56 0.16 0.15 0.70 0.32
The df2 stores item and its corresponding group information such as
df2 = pd.DataFrame({'item': item_names,
'group':['a1','a1','a1','a2',
'a2','a2','a2','a3','a3']})
df2 output:
item group
0 2 a1
1 7 a1
2 9 a1
3 10 a2
4 11 a2
5 13 a2
6 14 a2
7 21 a3
8 24 a3
The goal is to write a function which can select top N items in a specific row (item name) based on the corresponding values (largest) using these two DataFrames' information. However, the returned top N items and query item ALL MUST from 'different groups'. Such as
A query item (item = 10) is in the 4th row of df1 (item = 10). The top 2 returned items will be [9, 21] not [9, 14]. Since, item 10 is from group = a2 and any of returned items (top N) should not from a2 group. I have checked Scott Boston solution for a similar problem but it can't avoid the top N items and query item are from same group. Any suggestions? many thanks
IIUC, you want to select the N largest values excluding the values from the same group.
Here is a function that does this:
def get_top_N(idx, N=2):
group = df2.set_index('item')['group']
incl = group[group.ne(group[idx])].index
return df1.loc[idx, incl].nlargest(2).index.to_list()
get_top_N(10)
# [9, 21]
If you additionally want to ensure that all values are from different groups (this was unclear if a requirement, as this is the case for your example). You can additionally do:
def get_top_N_diff(idx, N=2):
group = df2.set_index('item')['group']
incl = group[group.ne(group[idx])].index
s = df1.loc[idx, incl]
return s.sort_values(ascending=False).groupby(group).idxmax().to_list()[:N]
get_top_N(11) # same group
# [9, 7]
get_top_N_diff(11) # different groups
# [9, 24]
Not sure exactly what you wanted... but this might point you in a direction:
import pandas as pd
import numpy as np
s2 = df2.set_index('item').group
mask = np.equal.outer(df1.index.map(s2.get), df1.columns.map(s2.get))
stacked = df1.mask(mask).stack().rename_axis(['x', 'y']).to_frame(name='v')
stacked.sort_values(['x', 'v'], ascending=[True, False]).groupby('x').head(2)
v
x y
2 14 0.98
11 0.72
7 14 0.74
10 0.44
9 10 0.85
11 0.72
10 9 0.63
21 0.43
11 9 0.94
7 0.89
13 9 0.99
21 0.60
14 24 0.62
21 0.59
21 7 0.84
10 0.76
24 7 0.72
2 0.63
A modification on the answer you mentioned:
def get_top(df1, df2, item_name, number_items):
val = df1.loc[[item_name]].T.merge(df2, left_index=True, right_on = 'item')
val = val[val['group']!=val.loc[val['item']==item_name, 'group'].values[0]]
return (val.sort_values(item_name, ascending=False)
.groupby('group')
.head(1)
.head(number_items)['item']
.to_numpy())
>>> get_top(df1, df2, 10, 2)
array([ 9, 21])
Related
I have a dataframe with multiple columns having numerical float values. What I want to do is give fractional weights to each column and calculate its average to store and append it to the same df.
Let's say we have the columns: s1, s2, s3
I want to give the weights: w1, w2, w3 to them respectively
I was able to do this manually while experimenting with all values in hand. But when I go to a list format, it's giving me an error.
I was trying to do it through iteration and I've attached my code below, but it was giving me an error. I have also attached my manual code which worked, but it needs it first hand.
Code which didn't work:
score_df["weighted_avg"] += weight * score_df[feature]
Manual Code which worked but not with lists:
df["weighted_scores"] = 0.5*df["s1"] + 0.25*df["s2"] + 0.25*df["s3"]
We can use numpy broadcasting for this, since weights has the same shape as your column axis:
# given the following example df
df = pd.DataFrame(np.random.rand(10,3), columns=["s1", "s2", "s3"])
print(df)
s1 s2 s3
0 0.49 1.00 0.50
1 0.65 0.87 0.75
2 0.45 0.85 0.87
3 0.91 0.53 0.30
4 0.96 0.44 0.50
5 0.67 0.87 0.24
6 0.87 0.41 0.29
7 0.06 0.15 0.73
8 0.76 0.92 0.69
9 0.92 0.28 0.29
weights = [0.5, 0.25, 0.25]
df["weighted_scores"] = df.mul(weights).sum(axis=1)
print(df)
s1 s2 s3 weighted_scores
0 0.49 1.00 0.50 0.62
1 0.65 0.87 0.75 0.73
2 0.45 0.85 0.87 0.66
3 0.91 0.53 0.30 0.66
4 0.96 0.44 0.50 0.71
5 0.67 0.87 0.24 0.61
6 0.87 0.41 0.29 0.61
7 0.06 0.15 0.73 0.25
8 0.76 0.92 0.69 0.78
9 0.92 0.28 0.29 0.60
You can use dot
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.rand(10,3), columns=["s1", "s2", "s3"])
df['weighted_scores'] = df.dot([.5,.25,.25])
df
Out
s1 s2 s3 weighted_scores
0 0.053543 0.659316 0.033540 0.199985
1 0.631627 0.257241 0.494959 0.503863
2 0.220939 0.870247 0.875165 0.546822
3 0.890487 0.519320 0.944459 0.811188
4 0.029416 0.016780 0.987503 0.265779
5 0.843882 0.784933 0.677096 0.787448
6 0.396092 0.297580 0.965454 0.513805
7 0.109894 0.011217 0.443796 0.168700
8 0.202096 0.637105 0.959876 0.500293
9 0.847020 0.949703 0.668615 0.828090
I have a pandas dataframe called df_initial with two columns 'a' and 'b' and N rows.
I would like to half the rows number, deleting the row where the value of 'b' is lower.
Thus between row 0 and row 1 I will keep row 1, between row 2 and row 3 I will keep row 3 etc..
This is the result that I would like to obtain:
print(df_initial)
a b
0 0.04 0.01
1 0.05 0.22
2 0.06 0.34
3 0.07 0.49
4 0.08 0.71
5 0.09 0.09
6 0.10 0.98
7 0.11 0.42
8 0.12 1.32
9 0.13 0.39
10 0.14 0.97
11 0.15 0.05
12 0.16 0.36
13 0.17 1.72
....
print(df_reduced)
a b
0 0.05 0.22
1 0.07 0.49
2 0.08 0.71
3 0.10 0.98
4 0.12 1.32
5 0.14 0.97
6 0.17 1.72
....
Is there some Pandas function to do this ?
I saw that there is a resample function, DataFrame.resample() , but it is valid with DatetimeIndex, TimedeltaIndex or PeriodIndex, so not in this case.
Thanks who will help me
You can groupby every two rows (a simple way of doing so is taking the floor division of the index) and take the idxmax of column b to index the dataframe:
df.loc[df.groupby(df.index//2).b.idxmax(), :]
a b
0 0.05 0.22
1 0.07 0.49
2 0.09 0.71
3 0.11 0.98
4 0.13 1.32
5 0.15 0.97
6 0.17 1.72
Or using DataFrame.rolling:
df.loc[df.b.rolling(2).max()[1::2].index, :]
This is an application for a simple example, you can apply it on your base.
import numpy as np
import pandas as pd
ar = np.array([[1.1, 1.0], [3.3, 0.2], [2.7, 10],[ 5.4, 7], [5.3, 9],[ 1.5, 15]])
df = pd.DataFrame(ar, columns = ['a', 'b'])
for i in range(len(df)):
if df['b'][i] < df['a'][i]:
df = df.drop(index = i)
print(df)````
I have a Pandas dataframe (Dt) like this:
Pc Cvt C1 C2 C3 C4 C5 C6 C7 C8 C9 C10
0 1 2 0.08 0.17 0.16 0.31 0.62 0.66 0.63 0.52 0.38
1 2 2 0.09 0.15 0.13 0.49 0.71 1.28 0.42 1.04 0.43
2 3 2 0.13 0.24 0.22 0.17 0.66 0.17 0.28 0.11 0.30
3 4 1 0.21 0.10 0.23 0.08 0.53 0.14 0.59 0.06 0.53
4 5 1 0.16 0.21 0.18 0.13 0.44 0.08 0.29 0.12 0.52
5 6 1 0.14 0.14 0.13 0.20 0.29 0.35 0.40 0.29 0.53
6 7 1 0.21 0.16 0.19 0.21 0.28 0.23 0.40 0.19 0.52
7 8 1 0.31 0.16 0.34 0.19 0.60 0.32 0.56 0.30 0.55
8 9 1 0.20 0.19 0.26 0.19 0.63 0.30 0.68 0.22 0.58
9 10 2 0.12 0.18 0.13 0.22 0.59 0.40 0.50 0.24 0.36
10 11 2 0.10 0.10 0.19 0.17 0.89 0.36 0.65 0.23 0.37
11 12 2 0.19 0.20 0.17 0.17 0.38 0.14 0.48 0.08 0.36
12 13 1 0.16 0.17 0.15 0.13 0.35 0.12 0.50 0.09 0.52
13 14 2 0.19 0.19 0.29 0.16 0.62 0.19 0.43 0.14 0.35
14 15 2 0.01 0.16 0.17 0.20 0.89 0.38 0.63 0.27 0.46
15 16 2 0.09 0.19 0.33 0.15 1.11 0.16 0.87 0.16 0.29
16 17 2 0.07 0.18 0.19 0.15 0.61 0.19 0.37 0.15 0.36
17 18 2 0.14 0.23 0.23 0.20 0.67 0.38 0.45 0.27 0.33
18 19 1 0.27 0.15 0.20 0.10 0.40 0.05 0.53 0.02 0.52
19 20 1 0.12 0.13 0.18 0.22 0.60 0.49 0.66 0.39 0.66
20 21 2 0.15 0.20 0.18 0.32 0.74 0.58 0.51 0.45 0.37
.
.
.
From this i want to plot an histogram with kde for each column from C1 to C10 in an arrange just like the one that i obtain if i plot it with pandas,
Dt.iloc[:,2:].hist()
But so far i've been not able to add the kde in each histogram; i want something like this:
Any ideas on how to accomplish this?
You want to first plot your histogram then plot the kde on a secondary axis.
Minimal and Complete Verifiable Example MCVE
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.DataFrame(np.random.randn(1000, 4)).add_prefix('C')
k = len(df.columns)
n = 2
m = (k - 1) // n + 1
fig, axes = plt.subplots(m, n, figsize=(n * 5, m * 3))
for i, (name, col) in enumerate(df.iteritems()):
r, c = i // n, i % n
ax = axes[r, c]
col.hist(ax=ax)
ax2 = col.plot.kde(ax=ax, secondary_y=True, title=name)
ax2.set_ylim(0)
fig.tight_layout()
How It Works
Keep track of total number of subplots
k = len(df.columns)
n will be the number of chart columns. Change this to suit individual needs. m will be the calculated number of required rows based on k and n
n = 2
m = (k - 1) // n + 1
Create a figure and array of axes with required number of rows and columns.
fig, axes = plt.subplots(m, n, figsize=(n * 5, m * 3))
Iterate through columns, tracking the column name and which number we are at i. Within each iteration, plot.
for i, (name, col) in enumerate(df.iteritems()):
r, c = i // n, i % n
ax = axes[r, c]
col.hist(ax=ax)
ax2 = col.plot.kde(ax=ax, secondary_y=True, title=name)
ax2.set_ylim(0)
Use tight_layout() as an easy way to sharpen up the layout spacing
fig.tight_layout()
Here is a pure seaborn solution, using FacetGrid.map_dataframe as explained here.
Stealing the example from #piRSquared:
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(1000, 4)).add_prefix('C')
Get the data in the required format:
df = df.stack().reset_index(level=1, name="val")
Result:
level_1 val
0 C0 0.879714
0 C1 -0.927096
0 C2 -0.929429
0 C3 -0.571176
1 C0 -1.127939
Then:
import seaborn as sns
def distplot(x, **kwargs):
ax = plt.gca()
data = kwargs.pop("data")
sns.distplot(data[x], ax=ax, **kwargs)
g = sns.FacetGrid(df, col="level_1", col_wrap=2, size=3.5)
g = g.map_dataframe(distplot, "val")
You can adjust col_wrap as needed.
I'm working with a frame like
df = pd.DataFrame({
'G1':[1.00,0.69,0.23,0.22,0.62],
'G2':[0.03,0.41,0.74,0.35,0.62],
'G3':[0.05,0.40,0.15,0.32,0.19],
'G4':[0.30,0.20,0.51,0.70,0.67],
'G5':[0.40,0.36,0.88,0.10,0.19]
})
and I want to manipulate it so that the columns are pairwise permutations of the current columns e.g. all columns are now 10 elements long and for example column 'G1:G2' would have column 'G2' appended to column 'G1'. I have attached a mock-up pic. Note that the pic has named indices unlike the above example code. I can work with or without the indices.
How could I approach this? I can make a function to act on each column, but I think the function would have to return a data frame made by concatenation with all other columns. Not sure what that would look like.
I'd do it like this
from itertools import permutations
l1, l2 = map(list, zip(*permutations(range(len(df.columns)), 2)))
v = df.values
pd.DataFrame(
np.vstack([v[:, l1], v[:, l2]]),
list(map('S{}'.format, range(1, len(df) + 1))) * 2,
df.columns.values[l1] + ':' + df.columns.values[l2]
)
Here is one way, although I suspect there might also be a way to do this directly in pandas
from itertools import permutations
'''Get all the column permutations'''
lst = [x for x in permutations(df.columns, 2)]
'''Create a list of columns names'''
names = [x[0]+'_'+x[1] for x in lst]
'''Create the new arrays by vertically stacking pairs of column values'''
cols = [np.vstack((df[x[0]].values,df[x[1]].values)).ravel() for x in lst]
'''Create a dictionary with column names as keys and the arrays as values'''
d = dict(zip(names, cols))
'''Create new dataframe from dict'''
df2 = pd.DataFrame(d)
df2
G1_G2 G1_G3 G1_G4 G1_G5 G2_G1 G2_G3 G2_G4 G2_G5 G3_G1 G3_G2 \
0 1.00 1.00 1.00 1.00 0.03 0.03 0.03 0.03 0.05 0.05
1 0.69 0.69 0.69 0.69 0.41 0.41 0.41 0.41 0.40 0.40
2 0.23 0.23 0.23 0.23 0.74 0.74 0.74 0.74 0.15 0.15
3 0.22 0.22 0.22 0.22 0.35 0.35 0.35 0.35 0.32 0.32
4 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.19 0.19
5 0.03 0.05 0.30 0.40 1.00 0.05 0.30 0.40 1.00 0.03
6 0.41 0.40 0.20 0.36 0.69 0.40 0.20 0.36 0.69 0.41
7 0.74 0.15 0.51 0.88 0.23 0.15 0.51 0.88 0.23 0.74
8 0.35 0.32 0.70 0.10 0.22 0.32 0.70 0.10 0.22 0.35
9 0.62 0.19 0.67 0.19 0.62 0.19 0.67 0.19 0.62 0.62
This is part of the output
To avoid creating the lists and use the fact that itertools.permutations is a generator:
d = dict((x[0]+'_'+x[1] , np.vstack((df[x[0]].values,df[x[1]].values)).ravel())
for x in permutations(df.columns, 2))
df2 = pd.DataFrame(d)
I have a pandas dataframe with about 100 columns of following type:
X1 Y1 X2 Y2 X3 Y3
0.78 0.22 0.19 0.42 0.04 0.65
0.43 0.29 0.43 0.84 0.14 0.42
0.57 0.70 0.59 0.86 0.11 0.40
0.92 0.52 0.81 0.33 0.54 1.00
w1here (X,Y) are basically pairs of values
I need to create the following from above.
X Y
0.78 0.22
0.43 0.29
0.57 0.70
0.92 0.52
0.19 0.42
0.43 0.84
0.59 0.86
0.81 0.33
0.04 0.65
0.14 0.42
0.11 0.40
0.54 1.00
i.e. stack all the X columns which are odd numbered and then stack all the Y columns which are even numbered.
I have no clue where to even start. For small number of columns I could easily have use the column names.
You can use lreshape, for column names use list comprehension:
x = [col for col in df.columns if 'X' in col]
y = [col for col in df.columns if 'Y' in col]
df = pd.lreshape(df, {'X': x,'Y': y})
print (df)
X Y
0 0.78 0.22
1 0.43 0.29
2 0.57 0.70
3 0.92 0.52
4 0.19 0.42
5 0.43 0.84
6 0.59 0.86
7 0.81 0.33
8 0.04 0.65
9 0.14 0.42
10 0.11 0.40
11 0.54 1.00
Solution with MultiIndex and stack:
df.columns = [np.arange(len(df.columns)) % 2, np.arange(len(df.columns)) // 2]
df = df.stack().reset_index(drop=True)
df.columns = ['X','Y']
print (df)
X Y
0 0.78 0.22
1 0.19 0.42
2 0.04 0.65
3 0.43 0.29
4 0.43 0.84
5 0.14 0.42
6 0.57 0.70
7 0.59 0.86
8 0.11 0.40
9 0.92 0.52
10 0.81 0.33
11 0.54 1.00
It may also be worth noting that you could just construct a new DataFrame explicitly with the X-Y values. This will most likely be quicker, but it assumes that the X-Y column pairs are the entirety of your DataFrame.
pd.DataFrame(dict(X=df.values[:,::2].reshape(-1),
Y=df.values[:,1::2].reshape(-1)))
Demo
>>> pd.DataFrame(dict(X=df.values[:,::2].reshape(-1),
Y=df.values[:,1::2].reshape(-1)))
X Y
0 0.78 0.22
1 0.19 0.42
2 0.04 0.65
3 0.43 0.29
4 0.43 0.84
5 0.14 0.42
6 0.57 0.70
7 0.59 0.86
8 0.11 0.40
9 0.92 0.52
10 0.81 0.33
11 0.54 1.00
You can use the documented pd.wide_to_long but you will need to use a 'dummy' column to uniquely identify each row. You can drop this column later.
pd.wide_to_long(df.reset_index(),
stubnames=['X', 'Y'],
i='index',
j='dropme').reset_index(drop=True)
X Y
0 0.78 0.22
1 0.43 0.29
2 0.57 0.70
3 0.92 0.52
4 0.19 0.42
5 0.43 0.84
6 0.59 0.86
7 0.81 0.33
8 0.04 0.65
9 0.14 0.42
10 0.11 0.40
11 0.54 1.00