Using a familiar Pandas for you:Reference Excel:In [95]: pd.read_excel(r'D:\temp\.data\a.xlsx') Out[95]: a b c 0 1 2 3 1 11 22 33 2 111 222 333 3 1111 2222 3333 Let's say we're interested in the first column.aIn [96]: pd.read_excel(r'D:\temp\.data\a.xlsx', usecols=[0]).T.to_csv(r'd:/temp/out.csv', header=None, index=False) out.csv:1,11,111,1111 Horse:In [97]: pd.read_excel(r'D:\temp\.data\a.xlsx', usecols=[0]) Out[97]: a 0 1 1 11 2 111 3 1111 In [98]: pd.read_excel(r'D:\temp.data\a.xlsx', usecols=[0]).T Out[98]: 0 1 2 3 a 1 11 111 1111

If I understood you correctly, I might need to. dataframe.merge()♪ Delete in one data frame a pole with zero, and then df1.merge(df2)♪

Number of cuts np.linspace(a, b, n) There must be a whole thing, so... n It's got to be a whole, not a shotgun, making it easier to do with a full-scale division. //:n = (b-a)//h + 1 In addition, the method .to_csv No parameter delimiterThere's a parameter. sep:df.to_csv('file.csv', index=False, sep=',') And commendable, of course, that you know how to combine the conditions in. PandasBut he understands. >=Two conditions don't need, one condition. In addition, .values No use when compared Series with number Pandas I understand that I'm being compared. valuesNot index:df = df[df.x >= b//2] P.S. The time-frame is fixed by the most elementary means itself. Pandas:df.plot(x='x', y='y');

Description code:data = [] for file in glob.glob('*.dat'): df = pd.read_csv(file, sep='\s+', engine='python', usecols=['RadVel'], nrows=2) data.append([file, df.loc[0, 'RadVel']]) res = pd.DataFrame(data, columns=['file', 'val']) res.to_csv('result.csv', sep='\t', index=False) NOTE: This code was not tested due to a lack of question https://ru.meta.stackoverflow.com/questions/7060/ ♪

If I understood the question correctly:grp = df.groupby("Датчик")["Показания"] first_idxs = grp.apply(lambda x: x.index[0]) df["счетчик"] = np.where( df["Показания"] != 0, grp.apply(lambda x: (x.index.isin(first_idxs) | x.eq(0)).cumsum()), 0 ) result:In [71]: df Out[71]: Дата Датчик Показания счетчик 0 01.01.2020 1 0 0 1 02.01.2020 1 2 1 2 03.01.2020 1 3 1 3 04.01.2020 1 5 1 4 05.01.2020 1 0 0 5 01.01.2020 1 22 2 6 02.01.2020 2 3 1 7 03.01.2020 2 0 0 8 04.01.2020 2 11 2 9 05.01.2020 2 22 2

X_train, X_test, y_train, y_test = train_test_split(... ^^^^^^^ ^^^^^^ ... (mean_absolute_error(train_y, pred_trainy), ^^^^^^^ mean_absolute_error(test_y, pred_testy))) ^^^^^^ Look carefully at the names of the variables I emphasized. Probably one of them. MAEYou stayed from other experiments and have other dimensions. Put it in the count. MAE variables y_train and y_test And everything will be fixed.

Try to use the Pandas by https://pandas.pydata.org/docs/reference/api/pandas.Series.mode.html :Dts[fnc].rolling(1000).apply(lambda x: x.mode()[0])

Something like that. If you only have these columns, you'll have an extra check. any I don't need it, especially since I clearly didn't write it properly.import pandas as pd df = pd.DataFrame({'int': [1, 2, 3], 'obj1': ['yes', 'no', 'yes'], 'obj2': ['no', 'yes', 'no'], 'obj3': ['n', 'y', 'n']}) obj_columns = df.columns[df.dtypes == 'object'].to_list() for col in obj_columns: if any(df[col].str.contains(x).any() for x in ('yes','no')): df[col] = df[col].map({'yes': 1, 'no': 0}) Out: int obj1 obj2 obj3 0 1 1 0 n 1 2 0 1 y 2 3 1 0 n

Try this:t = test_dict_df["name"].map(update_dict).dropna() test_dict_df.loc[t.index, "value"] = t result:In [276]: test_dict_df Out[276]: name value 0 A 1.0 1 B 2.0 2 C 3.0 3 D 33.0 PS method https://pandas.pydata.org/docs/reference/api/pandas.Series.map.html Replace all the keys with their respective values, so the cycle is not necessary.

First, UNPIVOT should be made on X* columns, transforming a table from broad to high, and then PIVOT using the X* columns as an index, and the Name/X columns as names of future columns:res = (df .set_index("Name/X") .stack() .reset_index() .pivot_table(index="level_1", columns="Name/X", values=0) .reset_index(drop=True) result:In [243]: res Out[243]: Name/X A B 0 10 22 1 12 14 2 33 44 3 31 17 And how to do the final table in the name/X column make no difference Line index and line number corresponding to number The column in the reference table?res = (df .set_index("Name/X") .stack() .reset_index() .pivot_table(index="level_1", columns="Name/X", values=0) .rename_axis(None)) res.index = res.index.str.replace(r"^\D+", "", regex=True) result:In [254]: res Out[254]: Name/X A B 1 10 22 2 12 14 3 33 44 4 31 17

Use it. idx = pd.IndexSlice slice_ = idx[:, "x1":"x3"] df.style.apply(highlight_max, props='color:white;background-color:green;', axis=1, subset=slice_)

Just like you said,df["new"] = df["brand"].str[0] + df["brand"].str[-1] + "-" + df.index.astype(str) result:In [22]: df Out[22]: ID brand model new 0 AI-0 AUDI TT AI-0

Working example using https://pandas.pydata.org/docs/reference/api/pandas.Series.str.extract.html :In [126]: df Out[126]: col 0 3/4 от 23.03.2021 1 1/670 от 20.07.2021 In [127]: df["dt"] = df["col"].str.extract(r"\b(\d{2}.\d{2}.\d{4})\b")[0] In [128]: df Out[128]: col dt 0 3/4 от 23.03.2021 23.03.2021 1 1/670 от 20.07.2021 20.07.2021 PS for the future - always on the question https://ru.meta.stackoverflow.com/questions/7060/ and the result you want.

I need a column installs less than a day after day. serve_time, date in column installs replaced by date in column server_time.Well, literally, as it says, in those lines that satisfy the condition, change this for this:condition = df['installs'] < df['server_time'] df.loc[condition, 'installs'] = df.loc[condition, 'server_time']

Group the poles. ["Код", "Название"] And make a pole. "Сумма":In [106]: df.groupby(["Код", "Название"])["Сумма"].sum().reset_index() Out[106]: Код Название Сумма 0 03.2.01.01-0001 Цемент 420 1 14.5.06.03-0002 Паста 187

Use the method. https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.dropna.html :res = df.dropna(thresh=int(len(df) * .8), axis=1) result:In [55]: res Out[55]: Col1 Col3 Col4 Col6 0 7 9.0 7 4 1 6 3.0 5 10 2 6 4.0 3 2 3 8 9.0 6 8 4 6 6.0 8 5 5 10 5.0 2 5 6 4 NaN 4 7 7 4 NaN 4 10 8 10 4.0 3 10 9 5 4.0 4 7 NOTE: You' df.dropna() removed only NaNnot gaps. If you have gaps in your column, this column will need to be converted first to the number of data:df["col_name"] = pd.to_numrtic(df["col_name"], errors="coerce")

Use the parameter. https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html :df = pd.read_csv("temp.csv", parse_dates=["Timestamp"]) result:In [114]: df.head() Out[114]: Timestamp F 0 2012-01-03 1 1 2012-01-06 1 2 2012-01-09 1 3 2012-01-12 1 4 2012-01-15 1 In [115]: df.dtypes Out[115]: Timestamp datetime64[ns] # <-- NOTE!!! F int64 dtype: object If there is at least one line in the data where time does not indicate midnight (00:00:00Then Pandas will show both date and time:In [116]: df = pd.read_csv("temp.csv", parse_dates=["Timestamp"]) In [117]: df.head() Out[117]: Timestamp F 0 2012-01-03 21:59:00 1 1 2012-01-06 00:00:00 1 2 2012-01-09 00:00:00 1 3 2012-01-12 00:00:00 1 4 2012-01-15 00:00:00 1 The time component can be cut as follows:In [118]: df["Timestamp"] = df["Timestamp"].dt.floor("D") In [119]: df.head() Out[119]: Timestamp F 0 2012-01-03 1 1 2012-01-06 1 2 2012-01-09 1 3 2012-01-12 1 4 2012-01-15 1 If you need to convert the date to a given format, it's done:df["timestamp_str"] = df["Timestamp"].dt.strftime("%m/%d/%Y") result:In [121]: df.head() Out[121]: Timestamp F timestamp_str 0 2012-01-03 1 01/03/2012 1 2012-01-06 1 01/06/2012 2 2012-01-09 1 01/09/2012 3 2012-01-12 1 01/12/2012 4 2012-01-15 1 01/15/2012 In [122]: df.dtypes Out[122]: Timestamp datetime64[ns] F int64 timestamp_str object # <-- NOTE!!! dtype: object NOTE: After this transformation, you will get the lines instead of the dates and, as a consequence, you will not be able to use any function to deal with dates/time, including arithmetic dates. Usually, dates are converted to a specific format only before they are retained in text form or for display/graph.

Working example:In [124]: np.random.seed(42) In [125]: df = pd.DataFrame(np.random.randint(20, size=(5, 3)), columns=list("abc")) In [126]: df Out[126]: a b c 0 6 19 14 1 10 7 6 2 18 10 10 3 3 7 2 4 1 11 5 In [127]: res = pd.DataFrame(np.sort(df, axis=0), columns=df.columns) In [128]: res Out[128]: a b c 0 1 7 2 1 3 7 5 2 6 10 6 3 10 11 10 4 18 19 14 NOTE: If you're sorted like this, you've got ties in your rows, that is, all columns will be sorted. independently of each other♪ The fact that before the grading belonged to one row may be scattered in different lines. That's the only way to realize what you described on the question.

The simplest thing in your case is to indicate the number of the column that should be an index when reading the file:data = pd.read_csv(filename, index_col=0) If it's important for you to have a name, you can do that:data = pd.read_csv(filename, index_col=0).rename_axis("ID компании") After that, you can rename the columns, but just take it that ID компании You don't have to tell me now that it's an index, not one of the pillars.

Use Pandas built-in techniques:In [265]: %timeit sr.rolling(1000).median() 14.4 ms ± 513 µs per loop (mean ± std. dev. of 7 runs, 100 loops each) In [266]: %timeit sr.rolling(1000).apply(lambda x: np.median(x)) 1.72 s ± 88.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) on my laptop, the difference is 119 times. ♪PS also questions - why do this many times? In the case of median values of different columns in one sliding window, this is also done by vectorized Pandas methods. free cycles♪