As you can see I have a dataframe with several columns with the same name but split into 0., 1. until 27.
How can I take all the values of 1.name and have it under 0.name?
Thank you very much!
Assuming that for all 0<=n<=27 the column names' suffixes are the same, one solution can be:
import pandas as pd
import re
# pattern to extract colum name suffix
pattern = re.compile('^\d\.([\w\.]+)')
# getting all the distinct column names/fields
fields = set([pattern.match(colname).group(1) for colname in df.columns])
# max prefix number, for you 27
n = 27
partitions = []
for i in range(0,n+1):
# creating column selector for partitions
columns_for_partition = list(map(lambda field: str(i) + f'.{field}', fields))
# get partition from dataframe and renaming column to field name (removing n. prefix)
partition = df[columns_for_partition].rename(lambda x: x.split('.',1)[1], axis=1)
partitions.append(partition)
new_df = pd.concat(partitions)
print(new_df)
With an initial dataframe df
0.name 0.something 1.name 1.something
0 a 1 d 4
1 b 2 e 5
2 c 3 f 6
The resulting dataframe new_df will look like:
name something
0 a 1
1 b 2
2 c 3
0 d 4
1 e 5
2 f 6
Related
The rows of clin.index (row length = 81) is a subset of the columns of common_mrna (col length = 151). I want to keep the columns of common_mrna only if the column names match to the row values of clin dataframe.
My code failed to reduce the number of columns in common_mrna to 81.
import pandas as pd
common_mrna = common_mrna.set_index("Hugo_Symbol")
mrna_val = {}
for colnames, val in common_mrna.iteritems():
for i, rows in clin.iterrows():
if [[common_mrna.columns == i] == "TRUE"]:
mrna_val = np.append(mrna_val, val)
mrna = np.concatenate(mrna_val, axis=0)
common_mrna
Hugo_Symbol
A
B
C
D
First
1
2
3
4
Second
5
row
6
7
clin
Another header
A
20
D
30
desired output
Hugo_Symbol
A
D
First
1
4
Second
5
7
Try this using reindex:
common_mrna.reindex(clin.index, axis=1)
Output:
A D
First 1 4
Second 5 7
Update, IIUC:
common_mrna.set_index('Hugo_Symbol').reindex(clin.index, axis=1).reset_index()
IUUC, you can select the rows of A header in clin found in common_mrna columns and add the first column of common_mrna
cols = clin.loc[clin.index.isin(common_mrna.columns)].index.tolist()
# or with set
cols = list(sorted(set(clin.index.tolist()) & set(common_mrna.columns), key=common_mrna.columns.tolist().index))
out = common_mrna[['Hugo_Symbol'] + cols]
print(out)
Hugo_Symbol A D
0 First 1 4
1 Second 5 7
I have the case where I want to sanity check labeled data. I have hundreds of features and want to find points which have the same features but different label. These found cluster of disagreeing labels should then be numbered and put into a new dataframe.
This isn't hard but I am wondering what the most elegant solution for this is.
Here an example:
import pandas as pd
df = pd.DataFrame({
"feature_1" : [0,0,0,4,4,2],
"feature_2" : [0,5,5,1,1,3],
"label" : ["A","A","B","B","D","A"]
})
result_df = pd.DataFrame({
"cluster_index" : [0,0,1,1],
"feature_1" : [0,0,4,4],
"feature_2" : [5,5,1,1],
"label" : ["A","B","B","D"]
})
In order to get the output you want (both de-duplication and cluster_index), you can use a groupby approach:
g = df.groupby(['feature_1', 'feature_2'])['label']
(df.assign(cluster_index=g.ngroup()) # get group name
.loc[g.transform('size').gt(1)] # filter the non-duplicates
# line below only to have a nice cluster_index range (0,1…)
.assign(cluster_index= lambda d: d['cluster_index'].factorize()[0])
)
output:
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
First get all duplicated values per feature columns and then if necessary remove duplciated by all columns (here in sample data not necessary), last add GroupBy.ngroup for groups indices:
df = df[df.duplicated(['feature_1','feature_2'],keep=False)].drop_duplicates()
df['cluster_index'] = df.groupby(['feature_1', 'feature_2'])['label'].ngroup()
print (df)
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
I am removing consecutive duplicates in groups in a dataframe. I am looking for a faster way than this:
def remove_consecutive_dupes(subdf):
dupe_ids = [ "A", "B" ]
is_duped = (subdf[dupe_ids].shift(-1) == subdf[dupe_ids]).all(axis=1)
subdf = subdf[~is_duped]
return subdf
# dataframe with columns key, A, B
df.groupby("key").apply(remove_consecutive_dupes).reset_index()
Is it possible to remove these without grouping first? Applying the above function to each group individually takes a lot of time, especially if the group count is like half the row count. Is there a way to do this operation on the entire dataframe at once?
A simple example for the algorithm if the above was not clear:
input:
key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5
5 x 1 2
output:
key A B
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
5 x 1 2
Row 2 was dropped because A=1 B=2 was also the previous row in group x.
Row 5 will not be dropped because it is not a consecutive duplicate in group x.
According to your code, you drop only lines if they appear below each other if
they are grouped by the key. So rows with another key inbetween do not influence this logic. But doing this, you want to preserve the original order of the records.
I guess the biggest influence in the runtime is the call of your function and
possibly not the grouping itself.
If you want to avoid this, you can try the following approach:
# create a column to restore the original order of the dataframe
df.reset_index(drop=True, inplace=True)
df.reset_index(drop=False, inplace=True)
df.columns= ['original_order'] + list(df.columns[1:])
# add a group column, that contains consecutive numbers if
# two consecutive rows differ in at least one of the columns
# key, A, B
compare_columns= ['key', 'A', 'B']
df.sort_values(['key', 'original_order'], inplace=True)
df['group']= (df[compare_columns] != df[compare_columns].shift(1)).any(axis=1).cumsum()
df.drop_duplicates(['group'], keep='first', inplace=True)
df.drop(columns=['group'], inplace=True)
# now just restore the original index and it's order
df.set_index('original_order', inplace=True)
df.sort_index(inplace=True)
df
Testing this, results in:
key A B
original_order
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
If you don't like the index name above (original_order), you just need to add the following line to remove it:
df.index.name= None
Testdata:
from io import StringIO
infile= StringIO(
""" key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5"""
)
df= pd.read_csv(infile, sep='\s+') #.set_index('Date')
df
I have 3 data frame:
df1
id,k,a,b,c
1,2,1,5,1
2,3,0,1,0
3,6,1,1,0
4,1,0,5,0
5,1,1,5,0
df2
name,a,b,c
p,4,6,8
q,1,2,3
df3
type,w_ave,vac,yak
n,3,5,6
v,2,1,4
from the multiplication, using pandas and numpy, I want to the output in df1:
id,k,a,b,c,w_ave,vac,yak
1,2,1,5,1,16,15,18
2,3,0,1,0,0,3,6
3,6,1,1,0,5,4,7
4,1,0,5,0,0,11,14
5,1,1,5,0,13,12,15
the conditions are:
The value of the new column will be =
#its not a code
df1["w_ave"][1] = df3["w_ave"]["v"]+ df1["a"][1]*df2["a"]["q"]+df1["b"][1]*df2["b"]["q"]+df1["c"][1]*df2["c"]["q"]
for output["w_ave"][1]= 2 +(1*1)+(5*2)+(1*3)
df3["w_ave"]["v"]=2
df1["a"][1]=1, df2["a"]["q"]=1 ;
df1["b"][1]=5, df2["b"]["q"]=2 ;
df1["c"][1]=1, df2["c"]["q"]=3 ;
Which means:
- a new column will be added in df1, from the name of the column from df3.
- for each row of the df1, the value of a, b, c will be multiplied with the same-named q value from df2. and summed together with the corresponding value of df3.
-the column name of df1 , matched will column name of df2 will be multiplied. The other not matched column will not be multiplied, like df1[k].
- However, if there is any 0 in df1["a"], the corresponding output will be zero.
I am struggling with this. It was tough to explain also. My attempts are very silly. I know this attempt will not work. However, I have added this:
import pandas as pd, numpy as np
data1 = "Sample_data1.csv"
data2 = "Sample_data2.csv"
data3 = "Sample_data3.csv"
folder = '~Sample_data/'
df1 =pd.read_csv(folder + data1)
df2 =pd.read_csv(folder + data2)
df3 =pd.read_csv(folder + data3)
df1= df2 * df1
Ok, so this will in no way resemble your desired output, but vectorizing the formula you provided:
df2=df2.set_index("name")
df3=df3.set_index("type")
df1["w_ave"] = df3.loc["v", "w_ave"]+ df1["a"].mul(df2.loc["q", "a"])+df1["b"].mul(df2.loc["q", "b"])+df1["c"].mul(df2.loc["q", "c"])
Outputs:
id k a b c w_ave
0 1 2 1 5 1 16
1 2 3 0 1 0 4
2 3 6 1 1 0 5
3 4 1 0 5 0 12
4 5 1 1 5 0 13
I have a pandas dataframe with column names like this:
id ColNameOrig_x ColNameOrig_y
There are many such columns, the 'x' and 'y' came about because 2 datasets with similar column names were merged.
What I need to do:
df.ColName = df.ColNameOrig_x + df.ColNameOrig_y
I am now manually repeating this line for many cols(close to 50), is there a wildcard way of doing this?
You can use DataFrame.filter with DataFrame.groupby by lambda function and axis=1 for grouping per columns names with aggregate sum or use text functions like Series.str.split with indexing:
df1 = df.filter(like='_').groupby(lambda x: x.split('_')[0], axis=1).sum()
print (df1)
ColName1Orig ColName2Orig
0 3 7
1 11 15
df1 = df.filter(like='_').groupby(df.columns.str.split('_').str[0], axis=1).sum()
print (df1)
ColName1Orig ColName2Orig
0 3 7
1 11 15
df1 = df.filter(like='_').groupby(df.columns.str[:12], axis=1).sum()
print (df1)
ColName1Orig ColName2Orig
0 3 7
1 11 15
You can use the subscripting syntax to access column names dynamically:
col_groups = ['ColName1', 'ColName2']
for grp in col_groups:
df[grp] = df[f'{grp}Orig_x'] + df[f'{grp}Orig_y']
Or you can aggregate by column group. For example
df = pd.DataFrame([
[1,2,3,4],
[5,6,7,8]
], columns=['ColName1Orig_x', 'ColName1Orig_y', 'ColName2Orig_x', 'ColName2Orig_y'])
# Here's your opportunity to define the wildcard
col_groups = df.columns.str.extract('(.+)Orig_[x|y]')[0]
df.columns = [col_groups, df.columns]
df.groupby(level=0, axis=1).sum()
Input:
ColName1Orig_x ColName1Orig_y ColName2Orig_x ColName2Orig_y
1 2 3 4
5 6 7 8
Output:
ColName1 ColName2
3 7
11 15