Python Pandas: Index is overlapped? How to fix

I have created a dataframe after importing weather data, now called "weather".

The end goal is to be able to view data for specific month and year.

It started like this: enter image description here

Then I ran weather = weather.T to transform the graph making it look like:
enter image description here

Then I ran weather.columns=weather.iloc[0] to make the graph look like:
enter image description here

But the "year" column and "month" column are located in the index (i think?). How would i get it so it looks like:
enter image description here

Thanks for looking! Will appreciate any help :)

Please note I will remove the first row with the years in it. So don't worry about this part

edited Nov 22 '18 at 0:03

jpp

102k2165116

asked Nov 21 '18 at 23:40

A Johnston

315

add a comment |

I have created a dataframe after importing weather data, now called "weather".

The end goal is to be able to view data for specific month and year.

It started like this: enter image description here

Then I ran weather = weather.T to transform the graph making it look like:
enter image description here

Then I ran weather.columns=weather.iloc[0] to make the graph look like:
enter image description here

But the "year" column and "month" column are located in the index (i think?). How would i get it so it looks like:
enter image description here

Thanks for looking! Will appreciate any help :)

Please note I will remove the first row with the years in it. So don't worry about this part

edited Nov 22 '18 at 0:03

jpp

102k2165116

asked Nov 21 '18 at 23:40

A Johnston

315

add a comment |

I have created a dataframe after importing weather data, now called "weather".

The end goal is to be able to view data for specific month and year.

It started like this: enter image description here

Then I ran weather = weather.T to transform the graph making it look like:
enter image description here

Then I ran weather.columns=weather.iloc[0] to make the graph look like:
enter image description here

But the "year" column and "month" column are located in the index (i think?). How would i get it so it looks like:
enter image description here

Thanks for looking! Will appreciate any help :)

Please note I will remove the first row with the years in it. So don't worry about this part

edited Nov 22 '18 at 0:03

jpp

102k2165116

asked Nov 21 '18 at 23:40

A Johnston

315

I have created a dataframe after importing weather data, now called "weather".

The end goal is to be able to view data for specific month and year.

It started like this: enter image description here

Then I ran weather = weather.T to transform the graph making it look like:
enter image description here

Then I ran weather.columns=weather.iloc[0] to make the graph look like:
enter image description here

But the "year" column and "month" column are located in the index (i think?). How would i get it so it looks like:
enter image description here

Thanks for looking! Will appreciate any help :)

Please note I will remove the first row with the years in it. So don't worry about this part

python pandas dataframe indexing

edited Nov 22 '18 at 0:03

jpp

102k2165116

asked Nov 21 '18 at 23:40

A Johnston

315

edited Nov 22 '18 at 0:03

jpp

102k2165116

asked Nov 21 '18 at 23:40

A Johnston

315

edited Nov 22 '18 at 0:03

jpp

102k2165116

edited Nov 22 '18 at 0:03

jpp

102k2165116

edited Nov 22 '18 at 0:03

jpp

102k2165116

asked Nov 21 '18 at 23:40

A Johnston

315

asked Nov 21 '18 at 23:40

A Johnston

315

asked Nov 21 '18 at 23:40

A Johnston

315

add a comment |

1 Answer
1

active

oldest

votes

This just means the pd.Index object underlying your pd.DataFrame object, unbeknownst to you, has a name:

df = pd.DataFrame({'YEAR': [2016, 2017, 2018],

                   'JAN': [1, 2, 3],

                   'FEB': [4, 5, 6],

                   'MAR': [7, 8, 9]})



df.columns.name = 'month'

df = df.T

df.columns = df.iloc[0]



print(df)



YEAR   2016  2017  2018

month                  

YEAR   2016  2017  2018

JAN       1     2     3

FEB       4     5     6

MAR       7     8     9

If this really bothers you, you can use reset_index to elevate your index to a series and then drop the extra heading row. You can, at the same time, remove the column name:

df = df.reset_index().drop(0)

df.columns.name = ''



print(df)



  month  2016  2017  2018

1   JAN     1     2     3

2   FEB     4     5     6

3   MAR     7     8     9

answered Nov 21 '18 at 23:47

jpp

102k2165116

@AJohnston, No problem, if this helped feel free to accept (tick on left) so other users know.

– jpp
Nov 21 '18 at 23:57

edit Thanks! That's how I planned on it looking :) To ask a follow up question.. how would you select the "5" value under "FEB" and "2017"? Thanks

– A Johnston
Nov 21 '18 at 23:58

df.loc[df['month'] == 'FEB', 2017].iat[0] might be sufficient, or df[2017].iat[1].

– jpp
Nov 21 '18 at 23:59

Thanks again! Another follow up if you don't mind How would I store that? Feb = weather.loc[weather['month'] == 'FEB', 2017].iat[0] Doesn't seem to work, any suggestions? Thanks

– A Johnston
Nov 22 '18 at 0:08

Sorry, not sure, what you're trying should work but if you're unable to figure it out after a while free free to post a new question.

– jpp
Nov 22 '18 at 0:09

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

This just means the pd.Index object underlying your pd.DataFrame object, unbeknownst to you, has a name:

df = pd.DataFrame({'YEAR': [2016, 2017, 2018],

                   'JAN': [1, 2, 3],

                   'FEB': [4, 5, 6],

                   'MAR': [7, 8, 9]})



df.columns.name = 'month'

df = df.T

df.columns = df.iloc[0]



print(df)



YEAR   2016  2017  2018

month                  

YEAR   2016  2017  2018

JAN       1     2     3

FEB       4     5     6

MAR       7     8     9

If this really bothers you, you can use reset_index to elevate your index to a series and then drop the extra heading row. You can, at the same time, remove the column name:

df = df.reset_index().drop(0)

df.columns.name = ''



print(df)



  month  2016  2017  2018

1   JAN     1     2     3

2   FEB     4     5     6

3   MAR     7     8     9

answered Nov 21 '18 at 23:47

jpp

102k2165116

@AJohnston, No problem, if this helped feel free to accept (tick on left) so other users know.

– jpp
Nov 21 '18 at 23:57

edit Thanks! That's how I planned on it looking :) To ask a follow up question.. how would you select the "5" value under "FEB" and "2017"? Thanks

– A Johnston
Nov 21 '18 at 23:58

df.loc[df['month'] == 'FEB', 2017].iat[0] might be sufficient, or df[2017].iat[1].

– jpp
Nov 21 '18 at 23:59

Thanks again! Another follow up if you don't mind How would I store that? Feb = weather.loc[weather['month'] == 'FEB', 2017].iat[0] Doesn't seem to work, any suggestions? Thanks

– A Johnston
Nov 22 '18 at 0:08

Sorry, not sure, what you're trying should work but if you're unable to figure it out after a while free free to post a new question.

– jpp
Nov 22 '18 at 0:09

add a comment |

This just means the pd.Index object underlying your pd.DataFrame object, unbeknownst to you, has a name:

df = pd.DataFrame({'YEAR': [2016, 2017, 2018],

                   'JAN': [1, 2, 3],

                   'FEB': [4, 5, 6],

                   'MAR': [7, 8, 9]})



df.columns.name = 'month'

df = df.T

df.columns = df.iloc[0]



print(df)



YEAR   2016  2017  2018

month                  

YEAR   2016  2017  2018

JAN       1     2     3

FEB       4     5     6

MAR       7     8     9

If this really bothers you, you can use reset_index to elevate your index to a series and then drop the extra heading row. You can, at the same time, remove the column name:

df = df.reset_index().drop(0)

df.columns.name = ''



print(df)



  month  2016  2017  2018

1   JAN     1     2     3

2   FEB     4     5     6

3   MAR     7     8     9

answered Nov 21 '18 at 23:47

jpp

102k2165116

@AJohnston, No problem, if this helped feel free to accept (tick on left) so other users know.

– jpp
Nov 21 '18 at 23:57

edit Thanks! That's how I planned on it looking :) To ask a follow up question.. how would you select the "5" value under "FEB" and "2017"? Thanks

– A Johnston
Nov 21 '18 at 23:58

df.loc[df['month'] == 'FEB', 2017].iat[0] might be sufficient, or df[2017].iat[1].

– jpp
Nov 21 '18 at 23:59

Thanks again! Another follow up if you don't mind How would I store that? Feb = weather.loc[weather['month'] == 'FEB', 2017].iat[0] Doesn't seem to work, any suggestions? Thanks

– A Johnston
Nov 22 '18 at 0:08

Sorry, not sure, what you're trying should work but if you're unable to figure it out after a while free free to post a new question.

– jpp
Nov 22 '18 at 0:09

add a comment |

This just means the pd.Index object underlying your pd.DataFrame object, unbeknownst to you, has a name:

df = pd.DataFrame({'YEAR': [2016, 2017, 2018],

                   'JAN': [1, 2, 3],

                   'FEB': [4, 5, 6],

                   'MAR': [7, 8, 9]})



df.columns.name = 'month'

df = df.T

df.columns = df.iloc[0]



print(df)



YEAR   2016  2017  2018

month                  

YEAR   2016  2017  2018

JAN       1     2     3

FEB       4     5     6

MAR       7     8     9

If this really bothers you, you can use reset_index to elevate your index to a series and then drop the extra heading row. You can, at the same time, remove the column name:

df = df.reset_index().drop(0)

df.columns.name = ''



print(df)



  month  2016  2017  2018

1   JAN     1     2     3

2   FEB     4     5     6

3   MAR     7     8     9

answered Nov 21 '18 at 23:47

jpp

102k2165116

This just means the pd.Index object underlying your pd.DataFrame object, unbeknownst to you, has a name:

df = pd.DataFrame({'YEAR': [2016, 2017, 2018],

                   'JAN': [1, 2, 3],

                   'FEB': [4, 5, 6],

                   'MAR': [7, 8, 9]})



df.columns.name = 'month'

df = df.T

df.columns = df.iloc[0]



print(df)



YEAR   2016  2017  2018

month                  

YEAR   2016  2017  2018

JAN       1     2     3

FEB       4     5     6

MAR       7     8     9

If this really bothers you, you can use reset_index to elevate your index to a series and then drop the extra heading row. You can, at the same time, remove the column name:

df = df.reset_index().drop(0)

df.columns.name = ''



print(df)



  month  2016  2017  2018

1   JAN     1     2     3

2   FEB     4     5     6

3   MAR     7     8     9

answered Nov 21 '18 at 23:47

jpp

102k2165116

answered Nov 21 '18 at 23:47

jpp

102k2165116

answered Nov 21 '18 at 23:47

jpp

102k2165116

answered Nov 21 '18 at 23:47

jpp

102k2165116

@AJohnston, No problem, if this helped feel free to accept (tick on left) so other users know.

– jpp
Nov 21 '18 at 23:57

edit Thanks! That's how I planned on it looking :) To ask a follow up question.. how would you select the "5" value under "FEB" and "2017"? Thanks

– A Johnston
Nov 21 '18 at 23:58

df.loc[df['month'] == 'FEB', 2017].iat[0] might be sufficient, or df[2017].iat[1].

– jpp
Nov 21 '18 at 23:59

Thanks again! Another follow up if you don't mind How would I store that? Feb = weather.loc[weather['month'] == 'FEB', 2017].iat[0] Doesn't seem to work, any suggestions? Thanks

– A Johnston
Nov 22 '18 at 0:08

Sorry, not sure, what you're trying should work but if you're unable to figure it out after a while free free to post a new question.

– jpp
Nov 22 '18 at 0:09

add a comment |

@AJohnston, No problem, if this helped feel free to accept (tick on left) so other users know.

– jpp
Nov 21 '18 at 23:57

edit Thanks! That's how I planned on it looking :) To ask a follow up question.. how would you select the "5" value under "FEB" and "2017"? Thanks

– A Johnston
Nov 21 '18 at 23:58

df.loc[df['month'] == 'FEB', 2017].iat[0] might be sufficient, or df[2017].iat[1].

– jpp
Nov 21 '18 at 23:59

Thanks again! Another follow up if you don't mind How would I store that? Feb = weather.loc[weather['month'] == 'FEB', 2017].iat[0] Doesn't seem to work, any suggestions? Thanks

– A Johnston
Nov 22 '18 at 0:08

Sorry, not sure, what you're trying should work but if you're unable to figure it out after a while free free to post a new question.

– jpp
Nov 22 '18 at 0:09

@AJohnston, No problem, if this helped feel free to accept (tick on left) so other users know.

– jpp
Nov 21 '18 at 23:57

edit Thanks! That's how I planned on it looking :) To ask a follow up question.. how would you select the "5" value under "FEB" and "2017"? Thanks

– A Johnston
Nov 21 '18 at 23:58

df.loc[df['month'] == 'FEB', 2017].iat[0] might be sufficient, or df[2017].iat[1].

– jpp
Nov 21 '18 at 23:59

Thanks again! Another follow up if you don't mind How would I store that? Feb = weather.loc[weather['month'] == 'FEB', 2017].iat[0] Doesn't seem to work, any suggestions? Thanks

– A Johnston
Nov 22 '18 at 0:08

Sorry, not sure, what you're trying should work but if you're unable to figure it out after a while free free to post a new question.

– jpp
Nov 22 '18 at 0:09

add a comment |

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