按名称将列移动到熊猫表的前面

Question 1

这是我的df：

                             Net   Upper   Lower  Mid  Zsore
Answer option                                                
More than once a day          0%   0.22%  -0.12%   2    65 
Once a day                    0%   0.32%  -0.19%   3    45
Several times a week          2%   2.45%   1.10%   4    78
Once a week                   1%   1.63%  -0.40%   6    65

如何将按名称（"Mid"）的列移动到表的前面，索引为0。结果应如下所示：

                             Mid   Upper   Lower  Net  Zsore
Answer option                                                
More than once a day          2   0.22%  -0.12%   0%    65 
Once a day                    3   0.32%  -0.19%   0%    45
Several times a week          4   2.45%   1.10%   2%    78
Once a week                   6   1.63%  -0.40%   1%    65

我当前的代码使用来按索引移动列，df.columns.tolist()但我想按名称进行移动。

Question 2

Here is my df:

                             Net   Upper   Lower  Mid  Zsore
Answer option                                                
More than once a day          0%   0.22%  -0.12%   2    65 
Once a day                    0%   0.32%  -0.19%   3    45
Several times a week          2%   2.45%   1.10%   4    78
Once a week                   1%   1.63%  -0.40%   6    65

How can I move a column by name ("Mid") to the front of the table, index 0. This is what the result should look like:

                             Mid   Upper   Lower  Net  Zsore
Answer option                                                
More than once a day          2   0.22%  -0.12%   0%    65 
Once a day                    3   0.32%  -0.19%   0%    45
Several times a week          4   2.45%   1.10%   2%    78
Once a week                   6   1.63%  -0.40%   1%    65

My current code moves the column by index using df.columns.tolist() but I’d like to shift it by name.

Question 3

我们可以ix通过传递列表来重新排序：

In [27]:
# get a list of columns
cols = list(df)
# move the column to head of list using index, pop and insert
cols.insert(0, cols.pop(cols.index('Mid')))
cols
Out[27]:
['Mid', 'Net', 'Upper', 'Lower', 'Zsore']
In [28]:
# use ix to reorder
df = df.ix[:, cols]
df
Out[28]:
                      Mid Net  Upper   Lower  Zsore
Answer_option                                      
More_than_once_a_day    2  0%  0.22%  -0.12%     65
Once_a_day              3  0%  0.32%  -0.19%     45
Several_times_a_week    4  2%  2.45%   1.10%     78
Once_a_week             6  1%  1.63%  -0.40%     65

另一种方法是引用该列，然后将其重新插入前面：

In [39]:
mid = df['Mid']
df.drop(labels=['Mid'], axis=1,inplace = True)
df.insert(0, 'Mid', mid)
df
Out[39]:
                      Mid Net  Upper   Lower  Zsore
Answer_option                                      
More_than_once_a_day    2  0%  0.22%  -0.12%     65
Once_a_day              3  0%  0.32%  -0.19%     45
Several_times_a_week    4  2%  2.45%   1.10%     78
Once_a_week             6  1%  1.63%  -0.40%     65

从以后开始，您还可以使用loc以获得与ix以后版本的熊猫不建议使用的相同的结果0.20.0：

df = df.loc[:, cols]

Question 4

We can use ix to reorder by passing a list:

In [27]:
# get a list of columns
cols = list(df)
# move the column to head of list using index, pop and insert
cols.insert(0, cols.pop(cols.index('Mid')))
cols
Out[27]:
['Mid', 'Net', 'Upper', 'Lower', 'Zsore']
In [28]:
# use ix to reorder
df = df.ix[:, cols]
df
Out[28]:
                      Mid Net  Upper   Lower  Zsore
Answer_option                                      
More_than_once_a_day    2  0%  0.22%  -0.12%     65
Once_a_day              3  0%  0.32%  -0.19%     45
Several_times_a_week    4  2%  2.45%   1.10%     78
Once_a_week             6  1%  1.63%  -0.40%     65

Another method is to take a reference to the column and reinsert it at the front:

In [39]:
mid = df['Mid']
df.drop(labels=['Mid'], axis=1,inplace = True)
df.insert(0, 'Mid', mid)
df
Out[39]:
                      Mid Net  Upper   Lower  Zsore
Answer_option                                      
More_than_once_a_day    2  0%  0.22%  -0.12%     65
Once_a_day              3  0%  0.32%  -0.19%     45
Several_times_a_week    4  2%  2.45%   1.10%     78
Once_a_week             6  1%  1.63%  -0.40%     65

You can also use loc to achieve the same result as ix will be deprecated in a future version of pandas from 0.20.0 onwards:

df = df.loc[:, cols]

Question 5

也许我错过了一些东西，但是许多答案似乎过于复杂。您应该只需要在一个列表中设置列即可：

列在最前面：

df = df[ ['Mid'] + [ col for col in df.columns if col != 'Mid' ] ]

或者，如果您想将其移到后面：

df = df[ [ col for col in df.columns if col != 'Mid' ] + ['Mid'] ]

或者，如果您想移动不止一列：

cols_to_move = ['Mid', 'Zsore']
df           = df[ cols_to_move + [ col for col in df.columns if col not in cols_to_move ] ]

Question 6

Maybe I’m missing something, but a lot of these answers seem overly complicated. You should be able to just set the columns within a single list:

Column to the front:

df = df[ ['Mid'] + [ col for col in df.columns if col != 'Mid' ] ]

Or if instead, you want to move it to the back:

df = df[ [ col for col in df.columns if col != 'Mid' ] + ['Mid'] ]

Or if you wanted to move more than one column:

cols_to_move = ['Mid', 'Zsore']
df           = df[ cols_to_move + [ col for col in df.columns if col not in cols_to_move ] ]

Question 7

您可以在熊猫中使用df.reindex（）函数。df是

                      Net  Upper   Lower  Mid  Zsore
Answer option                                      
More than once a day  0%  0.22%  -0.12%    2     65
Once a day            0%  0.32%  -0.19%    3     45
Several times a week  2%  2.45%   1.10%    4     78
Once a week           1%  1.63%  -0.40%    6     65

定义列名列表

cols = df.columns.tolist()
cols
Out[13]: ['Net', 'Upper', 'Lower', 'Mid', 'Zsore']

将列名移动到所需位置

cols.insert(0, cols.pop(cols.index('Mid')))
cols
Out[16]: ['Mid', 'Net', 'Upper', 'Lower', 'Zsore']

然后使用df.reindex()函数重新排序

df = df.reindex(columns= cols)

输出是：df

                      Mid  Upper   Lower Net  Zsore
Answer option                                      
More than once a day    2  0.22%  -0.12%  0%     65
Once a day              3  0.32%  -0.19%  0%     45
Several times a week    4  2.45%   1.10%  2%     78
Once a week             6  1.63%  -0.40%  1%     65

Question 8

You can use the df.reindex() function in pandas. df is

                      Net  Upper   Lower  Mid  Zsore
Answer option                                      
More than once a day  0%  0.22%  -0.12%    2     65
Once a day            0%  0.32%  -0.19%    3     45
Several times a week  2%  2.45%   1.10%    4     78
Once a week           1%  1.63%  -0.40%    6     65

define an list of column names

cols = df.columns.tolist()
cols
Out[13]: ['Net', 'Upper', 'Lower', 'Mid', 'Zsore']

move the column name to wherever you want

cols.insert(0, cols.pop(cols.index('Mid')))
cols
Out[16]: ['Mid', 'Net', 'Upper', 'Lower', 'Zsore']

then use df.reindex() function to reorder

df = df.reindex(columns= cols)

out put is: df

                      Mid  Upper   Lower Net  Zsore
Answer option                                      
More than once a day    2  0.22%  -0.12%  0%     65
Once a day              3  0.32%  -0.19%  0%     45
Several times a week    4  2.45%   1.10%  2%     78
Once a week             6  1.63%  -0.40%  1%     65

Question 9

我更喜欢这种解决方案：

col = df.pop("Mid")
df.insert(0, col.name, col)

它比其他建议的答案更容易阅读且速度更快。

def move_column_inplace(df, col, pos):
    col = df.pop(col)
    df.insert(pos, col.name, col)

绩效评估：

对于此测试，当前的最后一列在每次重复中都移到最前面。就地方法通常表现更好。虽然citynorman的解决方案可以就地完成，但Ed Chum的基于方法.loc和sachinnm的方法却reindex不能。

尽管其他方法通用，但citynorman的解决方案仅限于pos=0。我没有观察到df.loc[cols]和之间的性能差异df[cols]，这就是为什么我没有包含其他建议的原因。

我在MacBook Pro（2015年中）上使用python 3.6.8和pandas 0.24.2进行了测试。

import numpy as np
import pandas as pd

n_cols = 11
df = pd.DataFrame(np.random.randn(200000, n_cols),
                  columns=range(n_cols))

def move_column_inplace(df, col, pos):
    col = df.pop(col)
    df.insert(pos, col.name, col)

def move_to_front_normanius_inplace(df, col):
    move_column_inplace(df, col, 0)
    return df

def move_to_front_chum(df, col):
    cols = list(df)
    cols.insert(0, cols.pop(cols.index(col)))
    return df.loc[:, cols]

def move_to_front_chum_inplace(df, col):
    col = df[col]
    df.drop(col.name, axis=1, inplace=True)
    df.insert(0, col.name, col)
    return df

def move_to_front_elpastor(df, col):
    cols = [col] + [ c for c in df.columns if c!=col ]
    return df[cols] # or df.loc[cols]

def move_to_front_sachinmm(df, col):
    cols = df.columns.tolist()
    cols.insert(0, cols.pop(cols.index(col)))
    df = df.reindex(columns=cols, copy=False)
    return df

def move_to_front_citynorman_inplace(df, col):
    # This approach exploits that reset_index() moves the index
    # at the first position of the data frame.
    df.set_index(col, inplace=True)
    df.reset_index(inplace=True)
    return df

def test(method, df):
    col = np.random.randint(0, n_cols)
    method(df, col)

col = np.random.randint(0, n_cols)
ret_mine = move_to_front_normanius_inplace(df.copy(), col)
ret_chum1 = move_to_front_chum(df.copy(), col)
ret_chum2 = move_to_front_chum_inplace(df.copy(), col)
ret_elpas = move_to_front_elpastor(df.copy(), col)
ret_sach = move_to_front_sachinmm(df.copy(), col)
ret_city = move_to_front_citynorman_inplace(df.copy(), col)

# Assert equivalence of solutions.
assert(ret_mine.equals(ret_chum1))
assert(ret_mine.equals(ret_chum2))
assert(ret_mine.equals(ret_elpas))
assert(ret_mine.equals(ret_sach))
assert(ret_mine.equals(ret_city))

结果：

# For n_cols = 11:
%timeit test(move_to_front_normanius_inplace, df)
# 1.05 ms ± 42.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit test(move_to_front_citynorman_inplace, df)
# 1.68 ms ± 46.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit test(move_to_front_sachinmm, df)
# 3.24 ms ± 96.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum, df)
# 3.84 ms ± 114 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_elpastor, df)
# 3.85 ms ± 58.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum_inplace, df)
# 9.67 ms ± 101 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


# For n_cols = 31:
%timeit test(move_to_front_normanius_inplace, df)
# 1.26 ms ± 31.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_citynorman_inplace, df)
# 1.95 ms ± 260 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_sachinmm, df)
# 10.7 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum, df)
# 11.5 ms ± 869 µs per loop (mean ± std. dev. of 7 runs, 100 loops each
%timeit test(move_to_front_elpastor, df)
# 11.4 ms ± 598 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum_inplace, df)
# 31.4 ms ± 1.89 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Question 10

I prefer this solution:

col = df.pop("Mid")
df.insert(0, col.name, col)

It’s simpler to read and faster than other suggested answers.

def move_column_inplace(df, col, pos):
    col = df.pop(col)
    df.insert(pos, col.name, col)

Performance assessment:

For this test, the currently last column is moved to the front in each repetition. In-place methods generally perform better. While citynorman’s solution can be made in-place, Ed Chum’s method based on .loc and sachinnm’s method based on reindex cannot.

While other methods are generic, citynorman’s solution is limited to pos=0. I didn’t observe any performance difference between df.loc[cols] and df[cols], which is why I didn’t include some other suggestions.

I tested with python 3.6.8 and pandas 0.24.2 on a MacBook Pro (Mid 2015).

import numpy as np
import pandas as pd

n_cols = 11
df = pd.DataFrame(np.random.randn(200000, n_cols),
                  columns=range(n_cols))

def move_column_inplace(df, col, pos):
    col = df.pop(col)
    df.insert(pos, col.name, col)

def move_to_front_normanius_inplace(df, col):
    move_column_inplace(df, col, 0)
    return df

def move_to_front_chum(df, col):
    cols = list(df)
    cols.insert(0, cols.pop(cols.index(col)))
    return df.loc[:, cols]

def move_to_front_chum_inplace(df, col):
    col = df[col]
    df.drop(col.name, axis=1, inplace=True)
    df.insert(0, col.name, col)
    return df

def move_to_front_elpastor(df, col):
    cols = [col] + [ c for c in df.columns if c!=col ]
    return df[cols] # or df.loc[cols]

def move_to_front_sachinmm(df, col):
    cols = df.columns.tolist()
    cols.insert(0, cols.pop(cols.index(col)))
    df = df.reindex(columns=cols, copy=False)
    return df

def move_to_front_citynorman_inplace(df, col):
    # This approach exploits that reset_index() moves the index
    # at the first position of the data frame.
    df.set_index(col, inplace=True)
    df.reset_index(inplace=True)
    return df

def test(method, df):
    col = np.random.randint(0, n_cols)
    method(df, col)

col = np.random.randint(0, n_cols)
ret_mine = move_to_front_normanius_inplace(df.copy(), col)
ret_chum1 = move_to_front_chum(df.copy(), col)
ret_chum2 = move_to_front_chum_inplace(df.copy(), col)
ret_elpas = move_to_front_elpastor(df.copy(), col)
ret_sach = move_to_front_sachinmm(df.copy(), col)
ret_city = move_to_front_citynorman_inplace(df.copy(), col)

# Assert equivalence of solutions.
assert(ret_mine.equals(ret_chum1))
assert(ret_mine.equals(ret_chum2))
assert(ret_mine.equals(ret_elpas))
assert(ret_mine.equals(ret_sach))
assert(ret_mine.equals(ret_city))

Results:

# For n_cols = 11:
%timeit test(move_to_front_normanius_inplace, df)
# 1.05 ms ± 42.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit test(move_to_front_citynorman_inplace, df)
# 1.68 ms ± 46.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit test(move_to_front_sachinmm, df)
# 3.24 ms ± 96.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum, df)
# 3.84 ms ± 114 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_elpastor, df)
# 3.85 ms ± 58.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum_inplace, df)
# 9.67 ms ± 101 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


# For n_cols = 31:
%timeit test(move_to_front_normanius_inplace, df)
# 1.26 ms ± 31.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_citynorman_inplace, df)
# 1.95 ms ± 260 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_sachinmm, df)
# 10.7 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum, df)
# 11.5 ms ± 869 µs per loop (mean ± std. dev. of 7 runs, 100 loops each
%timeit test(move_to_front_elpastor, df)
# 11.4 ms ± 598 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum_inplace, df)
# 31.4 ms ± 1.89 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Question 11

我不喜欢必须在其他解决方案中明确指定所有其他列的方式，因此这对我来说最有效。虽然对于大型数据帧可能会比较慢…？

df = df.set_index('Mid').reset_index()

Question 12

I didn’t like how I had to explicitly specify all the other column in the other solutions so this worked best for me. Though it might be slow for large dataframes…?

df = df.set_index('Mid').reset_index()

Question 13

这是我经常使用的一组通用代码来重新排列列的位置。您可能会发现它很有用。

cols = df.columns.tolist()
n = int(cols.index('Mid'))
cols = [cols[n]] + cols[:n] + cols[n+1:]
df = df[cols]

Question 14

Here is a generic set of code that I frequently use to rearrange the position of columns. You may find it useful.

cols = df.columns.tolist()
n = int(cols.index('Mid'))
cols = [cols[n]] + cols[:n] + cols[n+1:]
df = df[cols]

Question 15

要重新排列DataFrame的行，只需使用如下列表即可。

df = df[['Mid', 'Net', 'Upper', 'Lower', 'Zsore']]

这使得以后阅读代码时所做的工作非常明显。也可以使用：

df.columns
Out[1]: Index(['Net', 'Upper', 'Lower', 'Mid', 'Zsore'], dtype='object')

然后剪切并粘贴以重新排序。

对于具有许多列的DataFrame，将列的列表存储在变量中，然后将所需的列弹出到列表的前面。这是一个例子：

cols = [str(col_name) for col_name in range(1001)]
data = np.random.rand(10,1001)
df = pd.DataFrame(data=data, columns=cols)

mv_col = cols.pop(cols.index('77'))
df = df[[mv_col] + cols]

现在df.columns有。

Index(['77', '0', '1', '2', '3', '4', '5', '6', '7', '8',
       ...
       '991', '992', '993', '994', '995', '996', '997', '998', '999', '1000'],
      dtype='object', length=1001)

Question 16

To reorder the rows of a DataFrame just use a list as follows.

df = df[['Mid', 'Net', 'Upper', 'Lower', 'Zsore']]

This makes it very obvious what was done when reading the code later. Also use:

df.columns
Out[1]: Index(['Net', 'Upper', 'Lower', 'Mid', 'Zsore'], dtype='object')

Then cut and paste to reorder.

For a DataFrame with many columns, store the list of columns in a variable and pop the desired column to the front of the list. Here is an example:

cols = [str(col_name) for col_name in range(1001)]
data = np.random.rand(10,1001)
df = pd.DataFrame(data=data, columns=cols)

mv_col = cols.pop(cols.index('77'))
df = df[[mv_col] + cols]

Now df.columns has.

Index(['77', '0', '1', '2', '3', '4', '5', '6', '7', '8',
       ...
       '991', '992', '993', '994', '995', '996', '997', '998', '999', '1000'],
      dtype='object', length=1001)

Question 17

这是一个非常简单的答案。

不要忘了在列名的两个（（））’括号’，否则会给你一个错误。


# here you can add below line and it should work 
df = df[list(('Mid','Upper', 'Lower', 'Net','Zsore'))]
df

                             Mid   Upper   Lower  Net  Zsore
Answer option                                                
More than once a day          2   0.22%  -0.12%   0%    65 
Once a day                    3   0.32%  -0.19%   0%    45
Several times a week          4   2.45%   1.10%   2%    78
Once a week                   6   1.63%  -0.40%   1%    65

Question 18

Here is a very simple answer to this.

Don’t forget the two (()) ‘brackets’ around columns names.Otherwise, it’ll give you an error.


# here you can add below line and it should work 
df = df[list(('Mid','Upper', 'Lower', 'Net','Zsore'))]
df

                             Mid   Upper   Lower  Net  Zsore
Answer option                                                
More than once a day          2   0.22%  -0.12%   0%    65 
Once a day                    3   0.32%  -0.19%   0%    45
Several times a week          4   2.45%   1.10%   2%    78
Once a week                   6   1.63%  -0.40%   1%    65

Question 19

您可以尝试的最简单的方法是：

df=df[[ 'Mid',   'Upper',   'Lower', 'Net'  , 'Zsore']]

Question 20

The most simplist thing you can try is:

df=df[[ 'Mid',   'Upper',   'Lower', 'Net'  , 'Zsore']]

按名称将列移动到熊猫表的前面

问题：按名称将列移动到熊猫表的前面

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回答 7

这是一个非常简单的答案。

Here is a very simple answer to this.

回答 8

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按名称将列移动到熊猫表的前面

问题：按名称将列移动到熊猫表的前面

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这是一个非常简单的答案。

Here is a very simple answer to this.

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