How to read all rows of a csv file using pandas in python?

Question

I am using the pandas module for reading the data from a .csv file.

I can write out the following code to extract the data belonging to an individual column as follows:

import pandas as pd

df = pd.read_csv('somefile.tsv', sep='\t', header=0)
some_column = df.column_name
print some_column # Gives the values of all entries in the column

However, the file that I am trying to read now has more than 5000 columns and writing out the statement

some_column = df.column_name

is now not feasible. How can I get all the column values so that I can access them using indexing?

eg to extract the value present at the 100th row and the 50th column, I should be able to write something like this:

df([100][50])

Answer 1

Use DataFrame.iloc or DataFrame.iat , but python counts from 0 , so need 99 and 49 for select 100. row and 50. column:

df = df.iloc[99,49]

Sample - select 3. row and 4. column:

df = pd.DataFrame({'A':[1,2,3],
                   'B':[4,5,6],
                   'C':[7,8,9],
                   'D':[1,3,10],
                   'E':[5,3,6],
                   'F':[7,4,3]})

print (df)
   A  B  C   D  E  F
0  1  4  7   1  5  7
1  2  5  8   3  3  4
2  3  6  9  10  6  3

print (df.iloc[2,3])
10

print (df.iat[2,3])
10

Combination for selecting by column name and position of row is possible by Series.iloc or Series.iat :

print (df['D'].iloc[2])
10

print (df['D'].iat[2])
10

Answer 2

Pandas has indexing for dataframes, so you can use

df.iloc[[index]]["column header"]

the index is in a list as you can pass multiple indexes at one in this way.