How to print a specific row of a pandas DataFrame?
Question:
I have a massive DataFrame, and I’m getting the error:
TypeError: ("Empty 'DataFrame': no numeric data to plot", 'occurred at index 159220')
I’ve already dropped nulls, and checked dtypes for the DataFrame so I have no guess as to why it’s failing on that row.
How do I print out just that row (at index 159220) of the DataFrame?
Answers:
Use ix operator:
print df.ix[159220]
Sounds like you’re calling df.plot(). That error indicates that you’re trying to plot a frame that has no numeric data. The data types shouldn’t affect what you print().
Use print(df.iloc[159220])
When you call loc with a scalar value, you get a pd.Series. That series will then have one dtype. If you want to see the row as it is in the dataframe, you’ll want to pass an array like indexer to loc.
Wrap your index value with an additional pair of square brackets
print(df.loc[[159220]])
To print a specific row, we have couple of pandas methods:
loc – It only gets the label i.e. column name or featuresiloc – Here i stands for integer, representing the row numberix – It is a mix of label as well as integer (not available in pandas >=1.0)
Below are examples of how to use the first two options for a specific row:
loc
df.loc[row,column]
For the first row and all columns:
df.loc[0,:]
For the first row and some specific column:
df.loc[0,'column_name']
iloc
For the first row and all columns:
df.iloc[0,:]
For the first row and some specific columns i.e. first three cols:
df.iloc[0,0:3]
If you want to display at row=159220
row=159220
#To display in a table format, you can use with or without display()
df.iloc[row:row+1]
df.loc[row:row] #Not recommended
#To display in print format, you can use with or without display()
df.iloc[row]
df.loc[row] #Not recommended
You can also index the index and use the result to select row(s) using loc:
row = 159220 # this creates a pandas Series (`row` is an integer)
row = [159220] # this creates a pandas DataFrame (`row` is a list)
df.loc[df.index[row]]
This is especially useful if you want to select rows by integer-location and columns by name. For example:
rows = 159220
cols = ['col2', 'col6']
df.loc[df.index[row], cols] # <--- OK
df.iloc[rows, cols] # <--- doesn't work
df.loc[cols].iloc[rows] # <--- OK but creates an intermediate copy
Print a specific row of a pandas DataFrame with example code with output:
Code:
import pandas as pd
data = {
'Name': ['Alice', 'Bob', 'Charlie', 'David', 'Emily'],
'Age': [25, 30, 22, 35, 28],
'Salary': [50000, 60000, 45000, 70000, 55000]
}
df = pd.DataFrame(data)
row_i_want_to_print = 3
specific_row = df.iloc[row_i_want_to_print-1] # here I want to print row=3 which index=2. so I did row_i_want_to_print-1 to print actual row
print("Selected Row: ")
print(specific_row)
Output:
Selected Row:
Name Charlie
Age 22
Salary 45000
Name: 2, dtype: object
I have a massive DataFrame, and I’m getting the error:
TypeError: ("Empty 'DataFrame': no numeric data to plot", 'occurred at index 159220')
I’ve already dropped nulls, and checked dtypes for the DataFrame so I have no guess as to why it’s failing on that row.
How do I print out just that row (at index 159220) of the DataFrame?
Use ix operator:
print df.ix[159220]
Sounds like you’re calling df.plot(). That error indicates that you’re trying to plot a frame that has no numeric data. The data types shouldn’t affect what you print().
Use print(df.iloc[159220])
When you call loc with a scalar value, you get a pd.Series. That series will then have one dtype. If you want to see the row as it is in the dataframe, you’ll want to pass an array like indexer to loc.
Wrap your index value with an additional pair of square brackets
print(df.loc[[159220]])
To print a specific row, we have couple of pandas methods:
loc– It only gets the label i.e. column name or featuresiloc– Here i stands for integer, representing the row numberix– It is a mix of label as well as integer (not available in pandas >=1.0)
Below are examples of how to use the first two options for a specific row:
loc
df.loc[row,column]
For the first row and all columns:
df.loc[0,:]
For the first row and some specific column:
df.loc[0,'column_name']
iloc
For the first row and all columns:
df.iloc[0,:]
For the first row and some specific columns i.e. first three cols:
df.iloc[0,0:3]
If you want to display at row=159220
row=159220
#To display in a table format, you can use with or without display()
df.iloc[row:row+1]
df.loc[row:row] #Not recommended
#To display in print format, you can use with or without display()
df.iloc[row]
df.loc[row] #Not recommended
You can also index the index and use the result to select row(s) using loc:
row = 159220 # this creates a pandas Series (`row` is an integer)
row = [159220] # this creates a pandas DataFrame (`row` is a list)
df.loc[df.index[row]]
This is especially useful if you want to select rows by integer-location and columns by name. For example:
rows = 159220
cols = ['col2', 'col6']
df.loc[df.index[row], cols] # <--- OK
df.iloc[rows, cols] # <--- doesn't work
df.loc[cols].iloc[rows] # <--- OK but creates an intermediate copy
Print a specific row of a pandas DataFrame with example code with output:
Code:
import pandas as pd
data = {
'Name': ['Alice', 'Bob', 'Charlie', 'David', 'Emily'],
'Age': [25, 30, 22, 35, 28],
'Salary': [50000, 60000, 45000, 70000, 55000]
}
df = pd.DataFrame(data)
row_i_want_to_print = 3
specific_row = df.iloc[row_i_want_to_print-1] # here I want to print row=3 which index=2. so I did row_i_want_to_print-1 to print actual row
print("Selected Row: ")
print(specific_row)
Output:
Selected Row:
Name Charlie
Age 22
Salary 45000
Name: 2, dtype: object