The column names (which are strings) cannot be sliced in the manner you tried.
Here you have a couple of options. If you know from context which variables you want to slice out, you can just return a view of only those columns by passing a list into the __getitem__ syntax (the []'s).
df1 = df[['a', 'b']]
Alternatively, if it matters to index them numerically and not by their name (say your code should automatically do this without knowing the names of the first two columns) then you can do this instead:
df1 = df.iloc[:, 0:2] # Remember that Python does not slice inclusive of the ending index.
Additionally, you should familiarize yourself with the idea of a view into a Pandas object vs. a copy of that object. The first of the above methods will return a new copy in memory of the desired sub-object (the desired slices).
Sometimes, however, there are indexing conventions in Pandas that don't do this and instead give you a new variable that just refers to the same chunk of memory as the sub-object or slice in the original object. This will happen with the second way of indexing, so you can modify it with the .copy() method to get a regular copy. When this happens, changing what you think is the sliced object can sometimes alter the original object. Always good to be on the look out for this.
df1 = df.iloc[0, 0:2].copy() # To avoid the case where changing df1 also changes df
To use iloc, you need to know the column positions (or indices). As the column positions may change, instead of hard-coding indices, you can use iloc along with get_loc function of columns method of dataframe object to obtain column indices.
{df.columns.get_loc(c): c for idx, c in enumerate(df.columns)}
Now you can use this dictionary to access columns through names and using iloc.
The column names (which are strings) cannot be sliced in the manner you tried.
Here you have a couple of options. If you know from context which variables you want to slice out, you can just return a view of only those columns by passing a list into the __getitem__ syntax (the []'s).
df1 = df[['a', 'b']]
Alternatively, if it matters to index them numerically and not by their name (say your code should automatically do this without knowing the names of the first two columns) then you can do this instead:
df1 = df.iloc[:, 0:2] # Remember that Python does not slice inclusive of the ending index.
Additionally, you should familiarize yourself with the idea of a view into a Pandas object vs. a copy of that object. The first of the above methods will return a new copy in memory of the desired sub-object (the desired slices).
Sometimes, however, there are indexing conventions in Pandas that don't do this and instead give you a new variable that just refers to the same chunk of memory as the sub-object or slice in the original object. This will happen with the second way of indexing, so you can modify it with the .copy() method to get a regular copy. When this happens, changing what you think is the sliced object can sometimes alter the original object. Always good to be on the look out for this.
df1 = df.iloc[0, 0:2].copy() # To avoid the case where changing df1 also changes df
To use iloc, you need to know the column positions (or indices). As the column positions may change, instead of hard-coding indices, you can use iloc along with get_loc function of columns method of dataframe object to obtain column indices.
{df.columns.get_loc(c): c for idx, c in enumerate(df.columns)}
Now you can use this dictionary to access columns through names and using iloc.
As of version 0.11.0, columns can be sliced in the manner you tried using the .loc indexer:
df.loc[:, 'C':'E']
is equivalent to
df[['C', 'D', 'E']] # or df.loc[:, ['C', 'D', 'E']]
and returns columns C through E.
A demo on a randomly generated DataFrame:
import pandas as pd
import numpy as np
np.random.seed(5)
df = pd.DataFrame(np.random.randint(100, size=(100, 6)),
columns=list('ABCDEF'),
index=['R{}'.format(i) for i in range(100)])
df.head()
Out:
A B C D E F
R0 99 78 61 16 73 8
R1 62 27 30 80 7 76
R2 15 53 80 27 44 77
R3 75 65 47 30 84 86
R4 18 9 41 62 1 82
To get the columns from C to E (note that unlike integer slicing, E is included in the columns):
df.loc[:, 'C':'E']
Out:
C D E
R0 61 16 73
R1 30 80 7
R2 80 27 44
R3 47 30 84
R4 41 62 1
R5 5 58 0
...
The same works for selecting rows based on labels. Get the rows R6 to R10 from those columns:
df.loc['R6':'R10', 'C':'E']
Out:
C D E
R6 51 27 31
R7 83 19 18
R8 11 67 65
R9 78 27 29
R10 7 16 94
.loc also accepts a Boolean array so you can select the columns whose corresponding entry in the array is True. For example, df.columns.isin(list('BCD')) returns array([False, True, True, True, False, False], dtype=bool) - True if the column name is in the list ['B', 'C', 'D']; False, otherwise.
df.loc[:, df.columns.isin(list('BCD'))]
Out:
B C D
R0 78 61 16
R1 27 30 80
R2 53 80 27
R3 65 47 30
R4 9 41 62
R5 78 5 58
...
You can remove one []:
df_new = df[list]
Also better is use other name as list, e.g. L:
df_new = df[L]
It look like working, I try only simplify it:
L = []
for x in df.columns:
if not "_" in x[-3:]:
L.append(x)
print (L)
List comprehension:
print ([x for x in df.columns if not "_" in x[-3:]])
1. [] aka __getitem__()
The canonical way to select a list of columns from a dataframe is via [].
df = pd.DataFrame({'a': [1, 1, 1, 1], 'b': [2, 2, 1, 0], 'c': [3, 3, 1, 0]})
cols = ['a', 'b']
df1 = df[cols]
Note that all column labels in cols have to also be df (otherwise KeyError: "... not in index" will be raised).
One thing to note is that when you want to assign new columns to df1 as filtered above (e.g. df1['new'] = 0), a SettingWithCopyWarning will be raised. To silence it, explicitly make a new copy:
df1 = df[cols].copy()
2. Handle KeyError: "... not in index"
To ensure cols contains only column labels that are in df, you can call isin on the columns and then filter df.
cols = ['a', 'b', 'f']
df1 = df[cols] # <----- error
df1 = df.loc[:, df.columns.isin(cols)] # <----- OK
3. filter()
Another way to select a list of columns from a dataframe is via filter(). A nice thing about it is that it creates a copy (so no SettingWithCopyWarning) and only selects the column labels in cols that exist in the dataframe, so handles the KeyError under the hood.
cols = ['a', 'b', 'f']
df1 = df.filter(cols)
As can be seen from the output below, f in cols is ignored because it doesn't exist as a column label in df.
If you can list all of the column names you want (as in, the number isn't massive), you can do this:
selection = df.filter(regex=("Level.*|Name|Qualification"))
The | character in the regex means or, so that line will take any column that matches one of:
"Level.*""Name""Qualification"
Probably lots of ways to go but I would select them separately then use
pd.concat([selection1, selection2], axis=1)
Maybe some other ideas in the merging docs.