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 following DataFrame (result of the method unstack):
df = pd.DataFrame(np.arange(12).reshape(2, -1),
columns=pd.CategoricalIndex(['a', 'b', 'c', 'a', 'b', 'c']))
df looks like this:
a b c a b c
0 0 1 2 3 4 5
1 6 7 8 9 10 11
When I try to df.reset_index() I get the following error:
TypeError: cannot insert an item into a CategoricalIndex that is not already an existing category
To bypass this problem I want to convert the column's index from categorical to a normal one. What is the most straightforward way to do it? Maybe you have an idea of how to reset the index without index conversion. I have the following idea:
df.columns = list(df.columns)
Most general is converting columns to list:
df.columns = df.columns.tolist()
Or if possible, convert them to strings:
df.columns = df.columns.astype(str)
df = df.reset_index()
print (df)
index a b c a b c
0 0 0 1 2 3 4 5
1 1 6 7 8 9 10 11
Can I insert a column at a specific column index in pandas?
import pandas as pd
df = pd.DataFrame({'l':['a','b','c','d'], 'v':[1,2,1,2]})
df['n'] = 0
This will put column n as the last column of df, but isn't there a way to tell df to put n at the beginning?
see docs: http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.insert.html
using loc = 0 will insert at the beginning
df.insert(loc, column, value)
df = pd.DataFrame({'B': [1, 2, 3], 'C': [4, 5, 6]})
df
Out:
B C
0 1 4
1 2 5
2 3 6
idx = 0
new_col = [7, 8, 9] # can be a list, a Series, an array or a scalar
df.insert(loc=idx, column='A', value=new_col)
df
Out:
A B C
0 7 1 4
1 8 2 5
2 9 3 6
If you want a single value for all rows:
df.insert(0,'name_of_column','')
df['name_of_column'] = value
Edit:
You can also:
df.insert(0,'name_of_column',value)
df.insert(loc, column_name, value)
This will work if there is no other column with the same name. If a column, with your provided name already exists in the dataframe, it will raise a ValueError.
You can pass an optional parameter allow_duplicates with True value to create a new column with already existing column name.
Here is an example:
>>> df = pd.DataFrame({'b': [1, 2], 'c': [3,4]})
>>> df
b c
0 1 3
1 2 4
>>> df.insert(0, 'a', -1)
>>> df
a b c
0 -1 1 3
1 -1 2 4
>>> df.insert(0, 'a', -2)
Traceback (most recent call last):
File "", line 1, in
File "C:\Python39\lib\site-packages\pandas\core\frame.py", line 3760, in insert
self._mgr.insert(loc, column, value, allow_duplicates=allow_duplicates)
File "C:\Python39\lib\site-packages\pandas\core\internals\managers.py", line 1191, in insert
raise ValueError(f"cannot insert {item}, already exists")
ValueError: cannot insert a, already exists
>>> df.insert(0, 'a', -2, allow_duplicates = True)
>>> df
a a b c
0 -2 -1 1 3
1 -2 -1 2 4
You could try to extract columns as list, massage this as you want, and reindex your dataframe:
>>> cols = df.columns.tolist()
>>> cols = [cols[-1]]+cols[:-1] # or whatever change you need
>>> df.reindex(columns=cols)
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
EDIT: this can be done in one line ; however, this looks a bit ugly. Maybe some cleaner proposal may come...
>>> df.reindex(columns=['n']+df.columns[:-1].tolist())
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
Here is a very simple answer to this(only one line).
You can do that after you added the 'n' column into your df as follows.
import pandas as pd
df = pd.DataFrame({'l':['a','b','c','d'], 'v':[1,2,1,2]})
df['n'] = 0
df
l v n
0 a 1 0
1 b 2 0
2 c 1 0
3 d 2 0
# here you can add the below code and it should work.
df = df[list('nlv')]
df
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
However, if you have words in your columns names instead of letters. It should include two brackets around your column names.
import pandas as pd
df = pd.DataFrame({'Upper':['a','b','c','d'], 'Lower':[1,2,1,2]})
df['Net'] = 0
df['Mid'] = 2
df['Zsore'] = 2
df
Upper Lower Net Mid Zsore
0 a 1 0 2 2
1 b 2 0 2 2
2 c 1 0 2 2
3 d 2 0 2 2
# here you can add below line and it should work
df = df[list(('Mid','Upper', 'Lower', 'Net','Zsore'))]
df
Mid Upper Lower Net Zsore
0 2 a 1 0 2
1 2 b 2 0 2
2 2 c 1 0 2
3 2 d 2 0 2
A general 4-line routine
You can have the following 4-line routine whenever you want to create a new column and insert into a specific location loc.
df['new_column'] = ... #new column's definition
col = df.columns.tolist()
col.insert(loc, col.pop()) #loc is the column's index you want to insert into
df = df[col]
In your example, it is simple:
df['n'] = 0
col = df.columns.tolist()
col.insert(0, col.pop())
df = df[col]
I have a DataFrame like this
DataFrame({"key":["a","b","c","d","e"], "value": [5,4,3,2,1]})
I am mainly interested in row "a", "b" and "c". I want to merge everything else into an "others" row like this
key value
0 a 5
1 b 4
2 c 3
3 others 3
I wonder how can this be done.
First create a dataframe without d and e:
df2 = df[df.key.isin(["a","b","c"])]
Then find the value that you want the other column to have (using the sum function in this example):
val = df[~df["key"].isin(["a","b","c"])].sum()["value"]
Finally, append this column to the second df:
df2.append({"key":"others", "value":val},ignore_index=True)
df2 is now:
key value
0 a 5
1 b 4
2 c 3
3 others 3
I have found a way to do it. Not sure if it is the best way.
In [3]: key_map = {"a":"a", "b":"b", "c":"c"}
In [4]: data['key1'] = data['key'].map(lambda k: key_map.get(k, "others"))
In [5]: data.groupby("key1").sum()
Out[5]:
value
key1
a 5
b 4
c 3
others 3
We can use .idxmax to get the maximum value of a dataframe(df). My problem is that I have a df with several columns (more than 10), one of a column has identifiers of same value. I need to extract the identifiers with the maximum value:
>df
id value
a 0
b 1
b 1
c 0
c 2
c 1
Now, this is what I'd want:
>df
id value
a 0
b 1
c 2
I am trying to get it by using df.groupy(['id']), but it is a bit tricky:
df.groupby(["id"]).ix[df['value'].idxmax()]
Of course, that doesn't work. I fear that I am not on the right path, so I thought I'd ask you guys! Thanks!
Close! Groupby the id, then use the value column; return the max for each group.
In [14]: df.groupby('id')['value'].max()
Out[14]:
id
a 0
b 1
c 2
Name: value, dtype: int64
Op wants to provide these locations back to the frame, just create a transform and assign.
In [17]: df['max'] = df.groupby('id')['value'].transform(lambda x: x.max())
In [18]: df
Out[18]:
id value max
0 a 0 0
1 b 1 1
2 b 1 1
3 c 0 2
4 c 2 2
5 c 1 2