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Deleting DataFrame row in Pandas based on column value - python

I have the following DataFrame:
daysago line_race rating rw wrating
line_date
2007-03-31 62 11 56 1.000000 56.000000
2007-03-10 83 11 67 1.000000 67.000000
2007-02-10 111 9 66 1.000000 66.000000
2007-01-13 139 10 83 0.880678 73.096278
2006-12-23 160 10 88 0.793033 69.786942
2006-11-09 204 9 52 0.636655 33.106077
2006-10-22 222 8 66 0.581946 38.408408
2006-09-29 245 9 70 0.518825 36.317752
2006-09-16 258 11 68 0.486226 33.063381
2006-08-30 275 8 72 0.446667 32.160051
2006-02-11 475 5 65 0.164591 10.698423
2006-01-13 504 0 70 0.142409 9.968634
2006-01-02 515 0 64 0.134800 8.627219
2005-12-06 542 0 70 0.117803 8.246238
2005-11-29 549 0 70 0.113758 7.963072
2005-11-22 556 0 -1 0.109852 -0.109852
2005-11-01 577 0 -1 0.098919 -0.098919
2005-10-20 589 0 -1 0.093168 -0.093168
2005-09-27 612 0 -1 0.083063 -0.083063
2005-09-07 632 0 -1 0.075171 -0.075171
2005-06-12 719 0 69 0.048690 3.359623
2005-05-29 733 0 -1 0.045404 -0.045404
2005-05-02 760 0 -1 0.039679 -0.039679
2005-04-02 790 0 -1 0.034160 -0.034160
2005-03-13 810 0 -1 0.030915 -0.030915
2004-11-09 934 0 -1 0.016647 -0.016647
I need to remove the rows where line_race is equal to 0. What's the most efficient way to do this?

If I'm understanding correctly, it should be as simple as:
df = df[df.line_race != 0]

But for any future bypassers you could mention that df = df[df.line_race != 0] doesn't do anything when trying to filter for None/missing values.
Does work:
df = df[df.line_race != 0]
Doesn't do anything:
df = df[df.line_race != None]
Does work:
df = df[df.line_race.notnull()]

just to add another solution, particularly useful if you are using the new pandas assessors, other solutions will replace the original pandas and lose the assessors
df.drop(df.loc[df['line_race']==0].index, inplace=True)

If you want to delete rows based on multiple values of the column, you could use:
df[(df.line_race != 0) & (df.line_race != 10)]
To drop all rows with values 0 and 10 for line_race.

In case of multiple values and str dtype
I used the following to filter out given values in a col:
def filter_rows_by_values(df, col, values):
return df[~df[col].isin(values)]
Example:
In a DataFrame I want to remove rows which have values "b" and "c" in column "str"
df = pd.DataFrame({"str": ["a","a","a","a","b","b","c"], "other": [1,2,3,4,5,6,7]})
df
str other
0 a 1
1 a 2
2 a 3
3 a 4
4 b 5
5 b 6
6 c 7
filter_rows_by_values(df, "str", ["b","c"])
str other
0 a 1
1 a 2
2 a 3
3 a 4

Though the previous answer are almost similar to what I am going to do, but using the index method does not require using another indexing method .loc(). It can be done in a similar but precise manner as
df.drop(df.index[df['line_race'] == 0], inplace = True)

The best way to do this is with boolean masking:
In [56]: df
Out[56]:
line_date daysago line_race rating raw wrating
0 2007-03-31 62 11 56 1.000 56.000
1 2007-03-10 83 11 67 1.000 67.000
2 2007-02-10 111 9 66 1.000 66.000
3 2007-01-13 139 10 83 0.881 73.096
4 2006-12-23 160 10 88 0.793 69.787
5 2006-11-09 204 9 52 0.637 33.106
6 2006-10-22 222 8 66 0.582 38.408
7 2006-09-29 245 9 70 0.519 36.318
8 2006-09-16 258 11 68 0.486 33.063
9 2006-08-30 275 8 72 0.447 32.160
10 2006-02-11 475 5 65 0.165 10.698
11 2006-01-13 504 0 70 0.142 9.969
12 2006-01-02 515 0 64 0.135 8.627
13 2005-12-06 542 0 70 0.118 8.246
14 2005-11-29 549 0 70 0.114 7.963
15 2005-11-22 556 0 -1 0.110 -0.110
16 2005-11-01 577 0 -1 0.099 -0.099
17 2005-10-20 589 0 -1 0.093 -0.093
18 2005-09-27 612 0 -1 0.083 -0.083
19 2005-09-07 632 0 -1 0.075 -0.075
20 2005-06-12 719 0 69 0.049 3.360
21 2005-05-29 733 0 -1 0.045 -0.045
22 2005-05-02 760 0 -1 0.040 -0.040
23 2005-04-02 790 0 -1 0.034 -0.034
24 2005-03-13 810 0 -1 0.031 -0.031
25 2004-11-09 934 0 -1 0.017 -0.017
In [57]: df[df.line_race != 0]
Out[57]:
line_date daysago line_race rating raw wrating
0 2007-03-31 62 11 56 1.000 56.000
1 2007-03-10 83 11 67 1.000 67.000
2 2007-02-10 111 9 66 1.000 66.000
3 2007-01-13 139 10 83 0.881 73.096
4 2006-12-23 160 10 88 0.793 69.787
5 2006-11-09 204 9 52 0.637 33.106
6 2006-10-22 222 8 66 0.582 38.408
7 2006-09-29 245 9 70 0.519 36.318
8 2006-09-16 258 11 68 0.486 33.063
9 2006-08-30 275 8 72 0.447 32.160
10 2006-02-11 475 5 65 0.165 10.698
UPDATE: Now that pandas 0.13 is out, another way to do this is df.query('line_race != 0').

The given answer is correct nontheless as someone above said you can use df.query('line_race != 0') which depending on your problem is much faster. Highly recommend.

Another way of doing it. May not be the most efficient way as the code looks a bit more complex than the code mentioned in other answers, but still alternate way of doing the same thing.
df = df.drop(df[df['line_race']==0].index)

One of the efficient and pandaic way is using eq() method:
df[~df.line_race.eq(0)]

I compiled and run my code. This is accurate code. You can try it your own.
data = pd.read_excel('file.xlsx')
If you have any special character or space in column name you can write it in '' like in the given code:
data = data[data['expire/t'].notnull()]
print (date)
If there is just a single string column name without any space or special
character you can directly access it.
data = data[data.expire ! = 0]
print (date)

Adding one more way to do this.
df = df.query("line_race!=0")

There are various ways to achieve that. Will leave below various options, that one can use, depending on specificities of one's use case.
One will consider that OP's dataframe is stored in the variable df.
Option 1
For OP's case, considering that the only column with values 0 is the line_race, the following will do the work
df_new = df[df != 0].dropna()
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.000000 56.000000
1 2007-03-10 83 11.0 67 1.000000 67.000000
2 2007-02-10 111 9.0 66 1.000000 66.000000
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
However, as that is not always the case, would recommend checking the following options where one will specify the column name.
Option 2
tshauck's approach ends up being better than Option 1, because one is able to specify the column. There are, however, additional variations depending on how one wants to refer to the column:
For example, using the position in the dataframe
df_new = df[df[df.columns[2]] != 0]
Or by explicitly indicating the column as follows
df_new = df[df['line_race'] != 0]
One can also follow the same login but using a custom lambda function, such as
df_new = df[df.apply(lambda x: x['line_race'] != 0, axis=1)]
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.000000 56.000000
1 2007-03-10 83 11.0 67 1.000000 67.000000
2 2007-02-10 111 9.0 66 1.000000 66.000000
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
Option 3
Using pandas.Series.map and a custom lambda function
df_new = df['line_race'].map(lambda x: x != 0)
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.000000 56.000000
1 2007-03-10 83 11.0 67 1.000000 67.000000
2 2007-02-10 111 9.0 66 1.000000 66.000000
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
Option 4
Using pandas.DataFrame.drop as follows
df_new = df.drop(df[df['line_race'] == 0].index)
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.000000 56.000000
1 2007-03-10 83 11.0 67 1.000000 67.000000
2 2007-02-10 111 9.0 66 1.000000 66.000000
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
Option 5
Using pandas.DataFrame.query as follows
df_new = df.query('line_race != 0')
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.000000 56.000000
1 2007-03-10 83 11.0 67 1.000000 67.000000
2 2007-02-10 111 9.0 66 1.000000 66.000000
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
Option 6
Using pandas.DataFrame.drop and pandas.DataFrame.query as follows
df_new = df.drop(df.query('line_race == 0').index)
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.000000 56.000000
1 2007-03-10 83 11.0 67 1.000000 67.000000
2 2007-02-10 111 9.0 66 1.000000 66.000000
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
Option 7
If one doesn't have strong opinions on the output, one can use a vectorized approach with numpy.select
df_new = np.select([df != 0], [df], default=np.nan)
[Out]:
[['2007-03-31' 62 11.0 56 1.0 56.0]
['2007-03-10' 83 11.0 67 1.0 67.0]
['2007-02-10' 111 9.0 66 1.0 66.0]
['2007-01-13' 139 10.0 83 0.880678 73.096278]
['2006-12-23' 160 10.0 88 0.793033 69.786942]
['2006-11-09' 204 9.0 52 0.636655 33.106077]
['2006-10-22' 222 8.0 66 0.581946 38.408408]
['2006-09-29' 245 9.0 70 0.518825 36.317752]
['2006-09-16' 258 11.0 68 0.486226 33.063381]
['2006-08-30' 275 8.0 72 0.446667 32.160051]
['2006-02-11' 475 5.0 65 0.164591 10.698423]]
This can also be converted to a dataframe with
df_new = pd.DataFrame(df_new, columns=df.columns)
[Out]:
line_date daysago line_race rating rw wrating
0 2007-03-31 62 11.0 56 1.0 56.0
1 2007-03-10 83 11.0 67 1.0 67.0
2 2007-02-10 111 9.0 66 1.0 66.0
3 2007-01-13 139 10.0 83 0.880678 73.096278
4 2006-12-23 160 10.0 88 0.793033 69.786942
5 2006-11-09 204 9.0 52 0.636655 33.106077
6 2006-10-22 222 8.0 66 0.581946 38.408408
7 2006-09-29 245 9.0 70 0.518825 36.317752
8 2006-09-16 258 11.0 68 0.486226 33.063381
9 2006-08-30 275 8.0 72 0.446667 32.160051
10 2006-02-11 475 5.0 65 0.164591 10.698423
With regards to the most efficient solution, that would depend on how one wants to measure efficiency. Assuming that one wants to measure the time of execution, one way that one can go about doing it is with time.perf_counter().
If one measures the time of execution for all the options above, one gets the following
method time
0 Option 1 0.00000110000837594271
1 Option 2.1 0.00000139995245262980
2 Option 2.2 0.00000369996996596456
3 Option 2.3 0.00000160001218318939
4 Option 3 0.00000110000837594271
5 Option 4 0.00000120000913739204
6 Option 5 0.00000140001066029072
7 Option 6 0.00000159995397552848
8 Option 7 0.00000150001142174006
However, this might change depending on the dataframe one uses, on the requirements (such as hardware), and more.
Notes:
There are various suggestions on using inplace=True. Would suggest reading this: https://stackoverflow.com/a/59242208/7109869
There are also some people with strong opinions on .apply(). Would suggest reading this: When should I (not) want to use pandas apply() in my code?
If one has missing values, one might want to consider as well pandas.DataFrame.dropna. Using the option 2, it would be something like
df = df[df['line_race'] != 0].dropna()
There are additional ways to measure the time of execution, so I would recommend this thread: How do I get time of a Python program's execution?

Just adding another way for DataFrame expanded over all columns:
for column in df.columns:
df = df[df[column]!=0]
Example:
def z_score(data,count):
threshold=3
for column in data.columns:
mean = np.mean(data[column])
std = np.std(data[column])
for i in data[column]:
zscore = (i-mean)/std
if(np.abs(zscore)>threshold):
count=count+1
data = data[data[column]!=i]
return data,count

Just in case you need to delete the row, but the value can be in different columns.
In my case I was using percentages so I wanted to delete the rows which has a value 1 in any column, since that means that it's the 100%
for x in df:
df.drop(df.loc[df[x]==1].index, inplace=True)
Is not optimal if your df have too many columns.

so many options provided(or maybe i didnt pay much attention to it, sorry if its the case), but no one mentioned this:
we can use this notation in pandas: ~ (this gives us the inverse of the condition)
df = df[~df["line_race"] == 0]

It doesn't make much difference for simple example like this, but for complicated logic, I prefer to use drop() when deleting rows because it is more straightforward than using inverse logic. For example, delete rows where A=1 AND (B=2 OR C=3).
Here's a scalable syntax that is easy to understand and can handle complicated logic:
df.drop( df.query(" `line_race` == 0 ").index)

You can try using this:
df.drop(df[df.line_race != 0].index, inplace = True)
.

Related

I cannot read the following data set in python, since it does not deliver each value in a column:( it is a .dat file)
6 3 289
1 154.0 417.0 510 193 .30
2 172.0 334.0 130 195 0 65 .23
3 267.0 87.0 70 105 0 35 .32
4 148.0 433.0 58 116 0 58 .33
5 355.0 444.0 48 72 0 24 .23
6 38.0 152.0 11 22 0 11 .18
I tried the np.loadtxt and the pd.read_csv functions and neither of them worked.
Would appreciate if someone could help me

I have a dataframe df1
index A B C D E
0 0 92 84
1 1 98 49
2 2 49 68
3 3 0 58
4 4 91 95
5 5 47 56 52 25 58
6 6 86 71 34 39 40
7 7 80 78 0 86 12
8 8 0 8 30 88 42
9 9 69 83 7 65 60
10 10 93 39 10 90 45
and also this data frame df2
index C D E F
0 0 27 95 51 45
1 1 99 33 92 67
2 2 68 37 29 65
3 3 99 25 48 40
4 4 33 74 55 66
5 13 65 76 19 62
I wish to get to the following outcome when merging df1 and df2
index A B C D E F
0 0 92 84 27 95 51 45
1 1 98 49 99 33 92 67
2 2 49 68 68 37 29 65
3 3 0 58 99 25 48 40
4 4 91 95 33 74 55 66
5 5 47 56 52 25 58
6 6 86 71 34 39 40
7 7 80 78 0 86 12
8 8 0 8 30 88 42
9 9 69 83 7 65 60
10 10 93 39 10 90 45
11 13 65 76 19 62
However, I am keeping getting this when using pd. merge(),
df_total=df1.merge(df2,how="outer",on="index",suffixes=(None,"_"))
df_total.replace(to_replace=np.nan,value=" ", inplace=True)
df_total
index A B C D E C_ D_ E_ F
0 0 92 84 27 95 51 45
1 1 98 49 99 33 92 67
2 2 49 68 68 37 29 65
3 3 0 58 99 25 48 40
4 4 91 95 33 74 55 66
5 5 47 56 52 25 58
6 6 86 71 34 39 40
7 7 80 78 0 86 12
8 8 0 8 30 88 42
9 9 69 83 7 65 60
10 10 93 39 10 90 45
11 13 65 76 19 62
Is there a way to get the desirable outcome using pd.merge or similar function?
Thanks

You can use .combine_first():
# convert the empty cells ("") to NaNs
df1 = df1.replace("", np.nan)
df2 = df2.replace("", np.nan)
# set indices and combine the dataframes
df1 = df1.set_index("index")
print(df1.combine_first(df2.set_index("index")).reset_index().fillna(""))
Prints:
index A B C D E F
0 0 92.0 84.0 27.0 95.0 51.0 45.0
1 1 98.0 49.0 99.0 33.0 92.0 67.0
2 2 49.0 68.0 68.0 37.0 29.0 65.0
3 3 0.0 58.0 99.0 25.0 48.0 40.0
4 4 91.0 95.0 33.0 74.0 55.0 66.0
5 5 47.0 56.0 52.0 25.0 58.0
6 6 86.0 71.0 34.0 39.0 40.0
7 7 80.0 78.0 0.0 86.0 12.0
8 8 0.0 8.0 30.0 88.0 42.0
9 9 69.0 83.0 7.0 65.0 60.0
10 10 93.0 39.0 10.0 90.0 45.0
11 13 65.0 76.0 19.0 62.0

I have a dataframe with ID's of clients and their expenses for 2014-2018. What I want is to have the mean of the expenses per ID but only the years before a certain date can be taken into account when calculating the mean value (so column 'Date' dictates which columns can be taken into account for the mean).
Example: for index 0 (ID: 12), the date states '2016-03-08', then the mean should be taken from the columns 'y_2014' and 'y_2015', so then for this index, the mean is 111.0.
If the date is too early (e.g. somewhere in 2014 or earlier in this case), then NaN should be returned (see index 6 and 9).
Initial dataframe:
y_2014 y_2015 y_2016 y_2017 y_2018 Date ID
0 100.0 122.0 324 632 NaN 2016-03-08 12
1 120.0 159.0 54 452 541.0 2015-04-09 96
2 NaN 164.0 687 165 245.0 2016-02-15 20
3 180.0 421.0 512 184 953.0 2018-05-01 73
4 110.0 654.0 913 173 103.0 2017-08-04 84
5 130.0 NaN 754 124 207.0 2016-07-03 26
6 170.0 256.0 843 97 806.0 2013-02-04 87
7 140.0 754.0 95 101 541.0 2016-06-08 64
8 80.0 985.0 184 84 90.0 2019-03-05 11
9 96.0 65.0 127 130 421.0 2014-05-14 34
Desired output:
y_2014 y_2015 y_2016 y_2017 y_2018 Date ID mean
0 100.0 122.0 324 632 NaN 2016-03-08 12 111.0
1 120.0 159.0 54 452 541.0 2015-04-09 96 120.0
2 NaN 164.0 687 165 245.0 2016-02-15 20 164.0
3 180.0 421.0 512 184 953.0 2018-05-01 73 324.25
4 110.0 654.0 913 173 103.0 2017-08-04 84 559.0
5 130.0 NaN 754 124 207.0 2016-07-03 26 130.0
6 170.0 256.0 843 97 806.0 2013-02-04 87 NaN
7 140.0 754.0 95 101 541.0 2016-06-08 64 447
8 80.0 985.0 184 84 90.0 2019-03-05 11 284.6
9 96.0 65.0 127 130 421.0 2014-05-14 34 NaN
Tried code: -> I'm still working on it, as I don't really know how to start for this, I only uploaded the dataframe so far, probably something with the 'datetime'-package has to be done to get the desired dataframe?
import pandas as pd
import numpy as np
import datetime
df = pd.DataFrame({"ID": [12,96,20,73,84,26,87,64,11,34],
"y_2014": [100,120,np.nan,180,110,130,170,140,80,96],
"y_2015": [122,159,164,421,654,np.nan,256,754,985,65],
"y_2016": [324,54,687,512,913,754,843,95,184,127],
"y_2017": [632,452,165,184,173,124,97,101,84,130],
"y_2018": [np.nan,541,245,953,103,207,806,541,90,421],
"Date": ['2016-03-08', '2015-04-09', '2016-02-15', '2018-05-01', '2017-08-04',
'2016-07-03', '2013-02-04', '2016-06-08', '2019-03-05', '2014-05-14']})
print(df)

Due to your naming convention, one need to extract the years from column names for comparison purpose. Then you can mask the data and taking mean:
# the years from columns
data = df.filter(like='y_')
data_years = data.columns.str.extract('(\d+)')[0].astype(int)
# the years from Date
years = pd.to_datetime(df.Date).dt.year.values
df['mean'] = data.where(data_years<years[:,None]).mean(1)
Output:
y_2014 y_2015 y_2016 y_2017 y_2018 Date ID mean
0 100.0 122.0 324 632 NaN 2016-03-08 12 111.00
1 120.0 159.0 54 452 541.0 2015-04-09 96 120.00
2 NaN 164.0 687 165 245.0 2016-02-15 20 164.00
3 180.0 421.0 512 184 953.0 2018-05-01 73 324.25
4 110.0 654.0 913 173 103.0 2017-08-04 84 559.00
5 130.0 NaN 754 124 207.0 2016-07-03 26 130.00
6 170.0 256.0 843 97 806.0 2013-02-04 87 NaN
7 140.0 754.0 95 101 541.0 2016-06-08 64 447.00
8 80.0 985.0 184 84 90.0 2019-03-05 11 284.60
9 96.0 65.0 127 130 421.0 2014-05-14 34 NaN

one more answer:
import pandas as pd
import numpy as np

df = pd.DataFrame({"ID": [12,96,20,73,84,26,87,64,11,34],
"y_2014": [100,120,np.nan,180,110,130,170,140,80,96],
"y_2015": [122,159,164,421,654,np.nan,256,754,985,65],
"y_2016": [324,54,687,512,913,754,843,95,184,127],
"y_2017": [632,452,165,184,173,124,97,101,84,130],
"y_2018": [np.nan,541,245,953,103,207,806,541,90,421],
"Date": ['2016-03-08', '2015-04-09', '2016-02-15', '2018-05-01', '2017-08-04',
'2016-07-03', '2013-02-04', '2016-06-08', '2019-03-05', '2014-05-14']})
#Subset from original df to calculate mean
subset = df.loc[:,['y_2014', 'y_2015', 'y_2016', 'y_2017', 'y_2018']]
#an expense value is only available for the calculation of the mean when that year has passed, therefore 2015-01-01 is chosen for the 'y_2014' column in the subset etc. to check with the 'Date'-column
subset.columns = ['2015-01-01', '2016-01-01', '2017-01-01', '2018-01-01', '2019-01-01']

s = subset.columns[0:].values < df.Date.values[:,None]
t = s.astype(float)
t[t == 0] = np.nan
df['mean'] = (subset.iloc[:,0:]*t).mean(1)

print(df)
#Additionally: (gives the sum of expenses before a certain date in the 'Date'-column
df['sum'] = (subset.iloc[:,0:]*t).sum(1)

print(df)

From excel file i want to read each row and use it independently to process
here how the data looks like in excel file
12 32 45 67 89 54 23 56 78 98
34 76 34 89 34 3
76 34 54 12 43 78 56
76 56 45 23 43 45 67 76 67 8
87 9 9 0 89 90 6 89
23 90 90 32 23 34 56 9 56 87
23 56 34 3 5 8 7 6 98
32 23 34 6 65 78 67 87 89 87
12 23 34 32 43 67 45
343 76 56 7 8 9 4
but when i read it through pandas then the remaining columns are filled with NaN.
the data after reading from pandas looks like
0 12 32 45 67 89 54 23.0 56.0 78.0 98.0
1 34 76 34 89 34 3 NaN NaN NaN NaN
2 76 34 54 12 43 78 56.0 NaN NaN NaN
3 76 56 45 23 43 45 67.0 76.0 67.0 8.0
4 87 9 9 0 89 90 6.0 89.0 NaN NaN
5 23 90 90 32 23 34 56.0 9.0 56.0 87.0
6 23 56 34 3 5 8 7.0 6.0 98.0 NaN
7 32 23 34 6 65 78 67.0 87.0 89.0 87.0
8 12 23 34 32 43 67 45.0 NaN NaN NaN
9 343 76 56 7 8 9 4.0 5.0 8.0 68.0
Here it can be seen the remaining columns of each row is filled with NaN which i don't want.
Nor i wanted to replace it with some other value or drop the whole rows contains NaN .
How can i read columns of each row till the first occurence of NaN. ?
For eg.The second row in pandas is 34 76 34 89 34 3 NaN NaN NaN NaN
so my desired output will be that it reads only 34 76 34 89 34 3
My preference is pandas but if not possible then is their any other way of doing it like with some other libraries
Any resource or reference will be helpful?
Thanks

While calling the pd.read_excel function, try setting keep_default_na = False. This will avoid default NaN values while reading.

I'm trying to extract data from a list of dataframes and extract row ranges. Each dataframe might not have the same data, therefore I have a list of possible index ranges that I would like loc to loop over, i.e. from the code sample below, I might want CIN to LAN, but on another dataframe, the CIN row doesn't exist, so I will want DET to LAN or HOU to LAN.
so I was thinking putting them in a list and iterating over the list, i.e.
for df in dfs:
ranges=[[df.loc["CIN":"LAN"]], [df.loc["DET":"LAN"]]]
extracted ranges = (i for i in ranges)
I'm not sure how you would iterate over a list and feed into loc, or perhaps .query().
df1 stint g ab r h X2b X3b hr rbi sb cs bb \
year team
2007 CIN 6 379 745 101 203 35 2 36 125.0 10.0 1.0 105
DET 5 301 1062 162 283 54 4 37 144.0 24.0 7.0 97
HOU 4 311 926 109 218 47 6 14 77.0 10.0 4.0 60
LAN 11 413 1021 153 293 61 3 36 154.0 7.0 5.0 114
NYN 13 622 1854 240 509 101 3 61 243.0 22.0 4.0 174
SFN 5 482 1305 198 337 67 6 40 171.0 26.0 7.0 235
TEX 2 198 729 115 200 40 4 28 115.0 21.0 4.0 73
TOR 4 459 1408 187 378 96 2 58 223.0 4.0 2.0 190
df2 so ibb hbp sh sf gidp
year team
2008 DET 176.0 3.0 10.0 4.0 8.0 28.0
HOU 212.0 3.0 9.0 16.0 6.0 17.0
LAN 141.0 8.0 9.0 3.0 8.0 29.0
NYN 310.0 24.0 23.0 18.0 15.0 48.0
SFN 188.0 51.0 8.0 16.0 6.0 41.0
TEX 140.0 4.0 5.0 2.0 8.0 16.0
TOR 265.0 16.0 12.0 4.0 16.0 38.0

Here is a solution:
import pandas as pd
# Prepare a list of ranges
ranges = [('CIN','LAN'), ('DET','LAN')]
# Declare an empty list of data frames and a list with the existing data frames
df_ranges = []
df_list = [df1, df2]
# Loop over multi-indices
for i, idx_range in enumerate(ranges):
df = df_list[i]
row1, row2 = idx_range
df_ranges.append(df.loc[(slice(None), slice(row1, row2)),:])
# Print the extracted data
print('Extracted data:\n')
print(df_ranges)
Output:
[ stint g ab r h X2b X3b hr rbi sb cs bb
year team
2007 CIN 6 379 745 101 203 35 2 36 125 10 1 105
DET 5 301 1062 162 283 54 4 37 144 24 7 97
HOU 4 311 926 109 218 47 6 14 77 10 4 60
LAN 11 413 1021 153 293 61 3 36 154 7 5 114
so ibb hbp sh sf gidp
year team
2008 DET 176 3 10 4 8 28
HOU 212 3 9 16 6 17
LAN 141 8 9 3 8 29]