Get column-index from column-name in pandas? [duplicate] - pandas

Question

In R when you need to retrieve a column index based on the name of the column you could do
idx <- which(names(my_data)==my_colum_name)
Is there a way to do the same with pandas dataframes?

Answer 1

Sure, you can use .get_loc():
In [45]: df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
In [46]: df.columns
Out[46]: Index([apple, orange, pear], dtype=object)
In [47]: df.columns.get_loc("pear")
Out[47]: 2
although to be honest I don't often need this myself. Usually access by name does what I want it to (df["pear"], df[["apple", "orange"]], or maybe df.columns.isin(["orange", "pear"])), although I can definitely see cases where you'd want the index number.

Answer 2

Here is a solution through list comprehension. cols is the list of columns to get index for:
[df.columns.get_loc(c) for c in cols if c in df]

Answer 3

DSM's solution works, but if you wanted a direct equivalent to which you could do (df.columns == name).nonzero()

Answer 4

For returning multiple column indices, I recommend using the pandas.Index method get_indexer, if you have unique labels:
df = pd.DataFrame({"pear": [1, 2, 3], "apple": [2, 3, 4], "orange": [3, 4, 5]})
df.columns.get_indexer(['pear', 'apple'])
# Out: array([0, 1], dtype=int64)
If you have non-unique labels in the index (columns only support unique labels) get_indexer_for. It takes the same args as get_indexer:
df = pd.DataFrame(
{"pear": [1, 2, 3], "apple": [2, 3, 4], "orange": [3, 4, 5]},
index=[0, 1, 1])
df.index.get_indexer_for([0, 1])
# Out: array([0, 1, 2], dtype=int64)
Both methods also support non-exact indexing with, f.i. for float values taking the nearest value with a tolerance. If two indices have the same distance to the specified label or are duplicates, the index with the larger index value is selected:
df = pd.DataFrame(
{"pear": [1, 2, 3], "apple": [2, 3, 4], "orange": [3, 4, 5]},
index=[0, .9, 1.1])
df.index.get_indexer([0, 1])
# array([ 0, -1], dtype=int64)

Answer 5

When you might be looking to find multiple column matches, a vectorized solution using searchsorted method could be used. Thus, with df as the dataframe and query_cols as the column names to be searched for, an implementation would be -
def column_index(df, query_cols):
cols = df.columns.values
sidx = np.argsort(cols)
return sidx[np.searchsorted(cols,query_cols,sorter=sidx)]
Sample run -
In [162]: df
Out[162]:
apple banana pear orange peach
0 8 3 4 4 2
1 4 4 3 0 1
2 1 2 6 8 1
In [163]: column_index(df, ['peach', 'banana', 'apple'])
Out[163]: array([4, 1, 0])

Answer 6

Update: "Deprecated since version 0.25.0: Use np.asarray(..) or DataFrame.values() instead." pandas docs
In case you want the column name from the column location (the other way around to the OP question), you can use:
>>> df.columns.values()[location]
Using #DSM Example:
>>> df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
>>> df.columns
Index(['apple', 'orange', 'pear'], dtype='object')
>>> df.columns.values()[1]
'orange'
Other ways:
df.iloc[:,1].name
df.columns[location] #(thanks to #roobie-nuby for pointing that out in comments.)

Answer 7

To modify DSM's answer a bit, get_loc has some weird properties depending on the type of index in the current version of Pandas (1.1.5) so depending on your Index type you might get back an index, a mask, or a slice. This is somewhat frustrating for me because I don't want to modify the entire columns just to extract one variable's index. Much simpler is to avoid the function altogether:
list(df.columns).index('pear')
Very straightforward and probably fairly quick.

Answer 8

how about this:
df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
out = np.argwhere(df.columns.isin(['apple', 'orange'])).ravel()
print(out)
[1 2]

Answer 9

When the column might or might not exist, then the following (variant from above works.
ix = 'none'
try:
ix = list(df.columns).index('Col_X')
except ValueError as e:
ix = None
pass
if ix is None:
# do something

Answer 10

import random
def char_range(c1, c2): # question 7001144
for c in range(ord(c1), ord(c2)+1):
yield chr(c)
df = pd.DataFrame()
for c in char_range('a', 'z'):
df[f'{c}'] = random.sample(range(10), 3) # Random Data
rearranged = random.sample(range(26), 26) # Random Order
df = df.iloc[:, rearranged]
print(df.iloc[:,:15]) # 15 Col View
for col in df.columns: # List of indices and columns
print(str(df.columns.get_loc(col)) + '\t' + col)
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