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Fixes a small typo/error in the Pandas indexing exercise 7 hint where 'df.iloc[0:999]' was written instead of 'df.loc[0:999]' #489

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Fixes a small typo/error in the Pandas indexing exercise 7 hint where 'df.iloc[0:999]' was written instead of 'df.loc[0:999]' #489

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Fix typo in DataFrame indexing exercise 7 hint
crazymakibe Jul 11, 2026
19f8032
Merge pull request #1 from crazymakibe/crazymakibe-patch-1
crazymakibe Jul 11, 2026
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2 changes: 1 addition & 1 deletion notebooks/pandas/raw/ex_1.ipynb
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Expand Up @@ -511,7 +511,7 @@
"> `iloc` uses the Python stdlib indexing scheme, where the first element of the range is included and the last one excluded. \n",
"`loc`, meanwhile, indexes inclusively. \n",
"\n",
"> This is particularly confusing when the DataFrame index is a simple numerical list, e.g. `0,...,1000`. In this case `df.iloc[0:1000]` will return 1000 entries, while `df.loc[0:1000]` return 1001 of them! To get 1000 elements using `loc`, you will need to go one lower and ask for `df.iloc[0:999]`. "
"> This is particularly confusing when the DataFrame index is a simple numerical list, e.g. `0,...,1000`. In this case `df.iloc[0:1000]` will return 1000 entries, while `df.loc[0:1000]` return 1001 of them! To get 1000 elements using `loc`, you will need to go one lower and ask for `df.loc[0:999]`. "
]
},
{
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