Why is Python Pandas loc not returning a single item?
For an assignment I need to access entries in a CSV file by the column name and index.
I am using a for loop to get each index. Contrary to my expectations,
read.loc[[i], ["magType"]]
returns something like
this
instead of just "mb" which is what I would expect based on a pandas cheatsheet (linked below)
Why is this? And how can I get just the item (in this case "mb" without the magType and 0)?
import pandas as pd
read = pd.read_csv("earthquake_data.csv")
print(read.loc[[0], ["magType"]])
magType
0 mb
https://s3.amazonaws.com/assets.datacamp.com/blog_assets/PandasPythonForDataScience.pdf
python pandas
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
add a comment |
For an assignment I need to access entries in a CSV file by the column name and index.
I am using a for loop to get each index. Contrary to my expectations,
read.loc[[i], ["magType"]]
returns something like
this
instead of just "mb" which is what I would expect based on a pandas cheatsheet (linked below)
Why is this? And how can I get just the item (in this case "mb" without the magType and 0)?
import pandas as pd
read = pd.read_csv("earthquake_data.csv")
print(read.loc[[0], ["magType"]])
magType
0 mb
https://s3.amazonaws.com/assets.datacamp.com/blog_assets/PandasPythonForDataScience.pdf
python pandas
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
add a comment |
For an assignment I need to access entries in a CSV file by the column name and index.
I am using a for loop to get each index. Contrary to my expectations,
read.loc[[i], ["magType"]]
returns something like
this
instead of just "mb" which is what I would expect based on a pandas cheatsheet (linked below)
Why is this? And how can I get just the item (in this case "mb" without the magType and 0)?
import pandas as pd
read = pd.read_csv("earthquake_data.csv")
print(read.loc[[0], ["magType"]])
magType
0 mb
https://s3.amazonaws.com/assets.datacamp.com/blog_assets/PandasPythonForDataScience.pdf
python pandas
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
For an assignment I need to access entries in a CSV file by the column name and index.
I am using a for loop to get each index. Contrary to my expectations,
read.loc[[i], ["magType"]]
returns something like
this
instead of just "mb" which is what I would expect based on a pandas cheatsheet (linked below)
Why is this? And how can I get just the item (in this case "mb" without the magType and 0)?
import pandas as pd
read = pd.read_csv("earthquake_data.csv")
print(read.loc[[0], ["magType"]])
magType
0 mb
https://s3.amazonaws.com/assets.datacamp.com/blog_assets/PandasPythonForDataScience.pdf
python pandas
python pandas
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
asked Nov 25 '18 at 23:52
L. ChenL. Chen
32
32
add a comment |
add a comment |
1 Answer
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oldest
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read.loc[0, "magType"] returns the element at 0-th row and magType column - presumably the result you desire.
read.loc[[0], ["magType"]], on the other hand, returns a slice of the DataFrame, with columns from ["magType"], with all the rows whose indices are in [0]. (Compare read.loc[[1, 2, 3], ["magType", "magnitude"]] or read.loc[[0:10], ["magType", "magnitude"]] to see a general case of this.) Like any other DataFrame, it has indices and column names.
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
I'd also throw in that the canonical way to access a scalar by row/column labels isat, i.e.read.at[0, 'magType']. Be careful with the description "the0th row", alignment between integer position and label is only true when your index is a regularpd.RangeIndex.
– jpp
Nov 26 '18 at 2:01
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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read.loc[0, "magType"] returns the element at 0-th row and magType column - presumably the result you desire.
read.loc[[0], ["magType"]], on the other hand, returns a slice of the DataFrame, with columns from ["magType"], with all the rows whose indices are in [0]. (Compare read.loc[[1, 2, 3], ["magType", "magnitude"]] or read.loc[[0:10], ["magType", "magnitude"]] to see a general case of this.) Like any other DataFrame, it has indices and column names.
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
I'd also throw in that the canonical way to access a scalar by row/column labels isat, i.e.read.at[0, 'magType']. Be careful with the description "the0th row", alignment between integer position and label is only true when your index is a regularpd.RangeIndex.
– jpp
Nov 26 '18 at 2:01
add a comment |
read.loc[0, "magType"] returns the element at 0-th row and magType column - presumably the result you desire.
read.loc[[0], ["magType"]], on the other hand, returns a slice of the DataFrame, with columns from ["magType"], with all the rows whose indices are in [0]. (Compare read.loc[[1, 2, 3], ["magType", "magnitude"]] or read.loc[[0:10], ["magType", "magnitude"]] to see a general case of this.) Like any other DataFrame, it has indices and column names.
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
I'd also throw in that the canonical way to access a scalar by row/column labels isat, i.e.read.at[0, 'magType']. Be careful with the description "the0th row", alignment between integer position and label is only true when your index is a regularpd.RangeIndex.
– jpp
Nov 26 '18 at 2:01
add a comment |
read.loc[0, "magType"] returns the element at 0-th row and magType column - presumably the result you desire.
read.loc[[0], ["magType"]], on the other hand, returns a slice of the DataFrame, with columns from ["magType"], with all the rows whose indices are in [0]. (Compare read.loc[[1, 2, 3], ["magType", "magnitude"]] or read.loc[[0:10], ["magType", "magnitude"]] to see a general case of this.) Like any other DataFrame, it has indices and column names.
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
read.loc[0, "magType"] returns the element at 0-th row and magType column - presumably the result you desire.
read.loc[[0], ["magType"]], on the other hand, returns a slice of the DataFrame, with columns from ["magType"], with all the rows whose indices are in [0]. (Compare read.loc[[1, 2, 3], ["magType", "magnitude"]] or read.loc[[0:10], ["magType", "magnitude"]] to see a general case of this.) Like any other DataFrame, it has indices and column names.
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
answered Nov 25 '18 at 23:58
AmadanAmadan
130k13142193
130k13142193
I'd also throw in that the canonical way to access a scalar by row/column labels isat, i.e.read.at[0, 'magType']. Be careful with the description "the0th row", alignment between integer position and label is only true when your index is a regularpd.RangeIndex.
– jpp
Nov 26 '18 at 2:01
add a comment |
I'd also throw in that the canonical way to access a scalar by row/column labels isat, i.e.read.at[0, 'magType']. Be careful with the description "the0th row", alignment between integer position and label is only true when your index is a regularpd.RangeIndex.
– jpp
Nov 26 '18 at 2:01
I'd also throw in that the canonical way to access a scalar by row/column labels is
at, i.e. read.at[0, 'magType']. Be careful with the description "the 0th row", alignment between integer position and label is only true when your index is a regular pd.RangeIndex.– jpp
Nov 26 '18 at 2:01
I'd also throw in that the canonical way to access a scalar by row/column labels is
at, i.e. read.at[0, 'magType']. Be careful with the description "the 0th row", alignment between integer position and label is only true when your index is a regular pd.RangeIndex.– jpp
Nov 26 '18 at 2:01
add a comment |
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