blocks|key|142743|text|Python列是Pandas+Series，当您将它们拉出时，您可以调用x.tolist()+on将它们转换为DataFrame列表。或者，您可以使用list(x)对其进行转换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|142744|import+pandas+as+pd

data_dict+=+{'one':+pd.Series([1,+2,+3],+index=['a',+'b',+'c']),
+++++++++++++'two':+pd.Series([1,+2,+3,+4],+index=['a',+'b',+'c',+'d'])}

df+=+pd.DataFrame(data_dict)

print(f"DataFrame:\n{df}\n")
print(f"column+types:\n{df.dtypes}")

col_one_list+=+df['one'].tolist()

col_one_arr+=+df['one'].to_numpy()

print(f"\ncol_one_list:\n{col_one_list}\ntype:{type(col_one_list)}")
print(f"\ncol_one_arr:\n{col_one_arr}\ntype:{type(col_one_arr)}")|code-block|syntax|javascript|142745|输出：|142746|DataFrame:
+++one++two
a++1.0++++1
b++2.0++++2
c++3.0++++3
d++NaN++++4

column+types:
one++++float64
two++++++int64
dtype:+object

col_one_list:
[1.0,+2.0,+3.0,+nan]
type:<class+'list'>

col_one_arr:
[+1.++2.++3.+nan]
type:<class+'numpy.ndarray'>|142747|entityMap^0|10|A|23|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Pandas DataFrame columns are Pandas Series when you pull them out, which you can then call <code>x.tolist()</code> on to turn them into a Python list. Alternatively you cast it with <code>list(x)</code>.
<pre class="lang-py prettyprint-override"><code>import pandas as pd

data_dict = {'one': pd.Series([1, 2, 3], index=['a', 'b', 'c']),
 'two': pd.Series([1, 2, 3, 4], index=['a', 'b', 'c', 'd'])}

df = pd.DataFrame(data_dict)

print(f&quot;DataFrame:\n{df}\n&quot;)
print(f&quot;column types:\n{df.dtypes}&quot;)

col_one_list = df['one'].tolist()

col_one_arr = df['one'].to_numpy()

print(f&quot;\ncol_one_list:\n{col_one_list}\ntype:{type(col_one_list)}&quot;)
print(f&quot;\ncol_one_arr:\n{col_one_arr}\ntype:{type(col_one_arr)}&quot;)
</code></pre>
Output:
<pre class="lang-none prettyprint-override"><code>DataFrame:
 one two
a 1.0 1
b 2.0 2
c 3.0 3
d NaN 4

column types:
one float64
two int64
dtype: object

col_one_list:
[1.0, 2.0, 3.0, nan]
type:&lt;class 'list'&gt;

col_one_arr:
[ 1. 2. 3. nan]
type:&lt;class 'numpy.ndarray'&gt;
</code></pre>

blocks|key|142807|text|示例转换：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|142808|一个Panda列中的Numpy+Array+->+Panda+Data+Frame+->列表|142809|Numpy数组|142810|data+=+np.array([[10,20,30],+[20,30,60],+[30,60,90]])|code-block|syntax|javascript|142811|将numpy数组转换为Panda数据帧|142812|dataPd+=+pd.DataFrame(data+=+data)
++++
print(dataPd)
0+++1+++2
0++10++20++30
1++20++30++60
2++30++60++90|142813|将一个Panda列转换为列表|142814|pdToList+=+list(dataPd['2'])|offset|length|style|CODE|142815|entityMap^0|0|0|0|0|0|0|0|0|S|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@$S|16|T|17|U|V]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

<h2>Example conversion:</h2>
Numpy Array -&gt; Panda Data Frame -&gt; List from one Panda Column
Numpy Array
<pre><code>data = np.array([[10,20,30], [20,30,60], [30,60,90]])
</code></pre>
Convert numpy array into Panda data frame
<pre><code>dataPd = pd.DataFrame(data = data)
 
print(dataPd)
0 1 2
0 10 20 30
1 20 30 60
2 30 60 90
</code></pre>
<h3>Convert one Panda column to list</h3>
<code>pdToList = list(dataPd['2'])</code>

blocks|key|4649433|text|假设读取excel表格后数据帧的名称为df，获取一个空列表(例如dataList)，逐行迭代数据帧并将其附加到您的空列表中，如下所示-|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4649434|dataList+=+[]+#empty+list
for+index,+row+in+df.iterrows():+
++++mylist+=+[row.cluster,+row.load_date,+row.budget,+row.actual,+row.fixed_price]
++++dataList.append(mylist)|code-block|syntax|javascript|4649435|或,|4649436|dataList+=+[]+#empty+list
for+row+in+df.itertuples():+
++++mylist+=+[row.cluster,+row.load_date,+row.budget,+row.actual,+row.fixed_price]
++++dataList.append(mylist)|4649437|不会，如果您打印dataList，您将在dataList中以列表的形式获得每一行。|4649438|entityMap^0|J|2|W|8|0|0|0|0|8|8|K|8|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

Assuming the name of the dataframe after reading the excel sheet is <code>df</code>, take an empty list (e.g. <code>dataList</code>), iterate through the dataframe row by row and append to your empty list like-

<pre><code>dataList = [] #empty list
for index, row in df.iterrows(): 
 mylist = [row.cluster, row.load_date, row.budget, row.actual, row.fixed_price]
 dataList.append(mylist)
</code></pre>

Or,

<pre><code>dataList = [] #empty list
for row in df.itertuples(): 
 mylist = [row.cluster, row.load_date, row.budget, row.actual, row.fixed_price]
 dataList.append(mylist)
</code></pre>

No, if you print the <code>dataList</code>, you will get each rows as a list in the <code>dataList</code>.

blocks|key|4649466|text|由于这个问题引起了很多人的关注，而且有几种方法可以完成你的任务，让我来介绍几种选择。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4649467|顺便说一下，这些都是一行程序;)|4649468|从以下内容开始：|4649469|df
++cluster+load_date+budget+actual+fixed_price
0+++++++A++1/1/2014+++1000+++4000+++++++++++Y
1+++++++A++2/1/2014++12000++10000+++++++++++Y
2+++++++A++3/1/2014++36000+++2000+++++++++++Y
3+++++++B++4/1/2014++15000++10000+++++++++++N
4+++++++B++4/1/2014++12000++11500+++++++++++N
5+++++++B++4/1/2014++90000++11000+++++++++++N
6+++++++C++7/1/2014++22000++18000+++++++++++N
7+++++++C++8/1/2014++30000++28960+++++++++++N
8+++++++C++9/1/2014++53000++51200+++++++++++N|code-block|syntax|javascript|4649470|潜在操作概述：|4649471|ser_aggCol+(collapse+each+column+to+a+list)
cluster++++++++++[A,+A,+A,+B,+B,+B,+C,+C,+C]
load_date++++++[1/1/2014,+2/1/2014,+3/1/2...
budget+++++++++[1000,+12000,+36000,+15000...
actual+++++++++[4000,+10000,+2000,+10000,...
fixed_price++++++[Y,+Y,+Y,+N,+N,+N,+N,+N,+N]
dtype:+object


ser_aggRows+(collapse+each+row+to+a+list)
0+++++[A,+1/1/2014,+1000,+4000,+Y]
1++++[A,+2/1/2014,+12000,+10000...
2++++[A,+3/1/2014,+36000,+2000,+Y]
3++++[B,+4/1/2014,+15000,+10000...
4++++[B,+4/1/2014,+12000,+11500...
5++++[B,+4/1/2014,+90000,+11000...
6++++[C,+7/1/2014,+22000,+18000...
7++++[C,+8/1/2014,+30000,+28960...
8++++[C,+9/1/2014,+53000,+51200...
dtype:+object


df_gr+(here+you+get+lists+for+each+cluster)
+++++++++++++++++++++++++++++load_date+++++++++++++++++budget+++++++++++++++++actual+fixed_price
cluster+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
A++++++++[1/1/2014,+2/1/2014,+3/1/2...+++[1000,+12000,+36000]++++[4000,+10000,+2000]+++[Y,+Y,+Y]
B++++++++[4/1/2014,+4/1/2014,+4/1/2...++[15000,+12000,+90000]++[10000,+11500,+11000]+++[N,+N,+N]
C++++++++[7/1/2014,+8/1/2014,+9/1/2...++[22000,+30000,+53000]++[18000,+28960,+51200]+++[N,+N,+N]


a+list+of+separate+dataframes+for+each+cluster

df+for+cluster+A
++cluster+load_date+budget+actual+fixed_price
0+++++++A++1/1/2014+++1000+++4000+++++++++++Y
1+++++++A++2/1/2014++12000++10000+++++++++++Y
2+++++++A++3/1/2014++36000+++2000+++++++++++Y

df+for+cluster+B
++cluster+load_date+budget+actual+fixed_price
3+++++++B++4/1/2014++15000++10000+++++++++++N
4+++++++B++4/1/2014++12000++11500+++++++++++N
5+++++++B++4/1/2014++90000++11000+++++++++++N

df+for+cluster+C
++cluster+load_date+budget+actual+fixed_price
6+++++++C++7/1/2014++22000++18000+++++++++++N
7+++++++C++8/1/2014++30000++28960+++++++++++N
8+++++++C++9/1/2014++53000++51200+++++++++++N

just+the+values+of+column+load_date
0++++1/1/2014
1++++2/1/2014
2++++3/1/2014
3++++4/1/2014
4++++4/1/2014
5++++4/1/2014
6++++7/1/2014
7++++8/1/2014
8++++9/1/2014
Name:+load_date,+dtype:+object


just+the+values+of+column+number+2
0+++++1000
1++++12000
2++++36000
3++++15000
4++++12000
5++++90000
6++++22000
7++++30000
8++++53000
Name:+budget,+dtype:+object


just+the+values+of+row+number+7
cluster+++++++++++++++C
load_date++++++8/1/2014
budget++++++++++++30000
actual++++++++++++28960
fixed_price+++++++++++N
Name:+7,+dtype:+object


==============================+JUST+FOR+COMPLETENESS+==============================


you+can+convert+a+series+to+a+list
['C',+'8/1/2014',+'30000',+'28960',+'N']
<class+'list'>


you+can+convert+a+dataframe+to+a+nested+list
[['A',+'1/1/2014',+'1000',+'4000',+'Y'],+['A',+'2/1/2014',+'12000',+'10000',+'Y'],+['A',+'3/1/2014',+'36000',+'2000',+'Y'],+['B',+'4/1/2014',+'15000',+'10000',+'N'],+['B',+'4/1/2014',+'12000',+'11500',+'N'],+['B',+'4/1/2014',+'90000',+'11000',+'N'],+['C',+'7/1/2014',+'22000',+'18000',+'N'],+['C',+'8/1/2014',+'30000',+'28960',+'N'],+['C',+'9/1/2014',+'53000',+'51200',+'N']]
<class+'list'>

the+content+of+a+dataframe+can+be+accessed+as+a+numpy.ndarray
[['A'+'1/1/2014'+'1000'+'4000'+'Y']
+['A'+'2/1/2014'+'12000'+'10000'+'Y']
+['A'+'3/1/2014'+'36000'+'2000'+'Y']
+['B'+'4/1/2014'+'15000'+'10000'+'N']
+['B'+'4/1/2014'+'12000'+'11500'+'N']
+['B'+'4/1/2014'+'90000'+'11000'+'N']
+['C'+'7/1/2014'+'22000'+'18000'+'N']
+['C'+'8/1/2014'+'30000'+'28960'+'N']
+['C'+'9/1/2014'+'53000'+'51200'+'N']]
<class+'numpy.ndarray'>|4649472|代码：|4649473|#+prefix+ser+refers+to+pd.Series+object
#+prefix+df+refers+to+pd.DataFrame+object
#+prefix+lst+refers+to+list+object

import+pandas+as+pd
import+numpy+as+np

df=pd.DataFrame([
++++++++['A',+++'1/1/2014',++++'1000',++++'4000',++++'Y'],
++++++++['A',+++'2/1/2014',++++'12000',+++'10000',+++'Y'],
++++++++['A',+++'3/1/2014',++++'36000',+++'2000',++++'Y'],
++++++++['B',+++'4/1/2014',++++'15000',+++'10000',+++'N'],
++++++++['B',+++'4/1/2014',++++'12000',+++'11500',+++'N'],
++++++++['B',+++'4/1/2014',++++'90000',+++'11000',+++'N'],
++++++++['C',+++'7/1/2014',++++'22000',+++'18000',+++'N'],
++++++++['C',+++'8/1/2014',++++'30000',+++'28960',+++'N'],
++++++++['C',+++'9/1/2014',++++'53000',+++'51200',+++'N']
++++++++],+columns=['cluster',+'load_date',+++'budget',++'actual',++'fixed_price'])
print('df',df,+sep='\n',+end='\n\n')

ser_aggCol=df.aggregate(lambda+x:+[x.tolist()],+axis=0).map(lambda+x:x[0])
print('ser_aggCol+(collapse+each+column+to+a+list)',ser_aggCol,+sep='\n',+end='\n\n\n')

ser_aggRows=pd.Series(df.values.tolist())+
print('ser_aggRows+(collapse+each+row+to+a+list)',ser_aggRows,+sep='\n',+end='\n\n\n')

df_gr=df.groupby('cluster').agg(lambda+x:+list(x))
print('df_gr+(here+you+get+lists+for+each+cluster)',df_gr,+sep='\n',+end='\n\n\n')

lst_dfFiltGr=[+df.loc[df['cluster']==val,:]+for+val+in+df['cluster'].unique()+]
print('a+list+of+separate+dataframes+for+each+cluster',+sep='\n',+end='\n\n')
for+dfTmp+in+lst_dfFiltGr:
++++print('df+for+cluster+'%2Bstr(dfTmp.loc[dfTmp.index[0],'cluster']),dfTmp,+sep='\n',+end='\n\n')

ser_singleColLD=df.loc[:,'load_date']
print('just+the+values+of+column+load_date',ser_singleColLD,+sep='\n',+end='\n\n\n')

ser_singleCol2=df.iloc[:,2]
print('just+the+values+of+column+number+2',ser_singleCol2,+sep='\n',+end='\n\n\n')

ser_singleRow7=df.iloc[7,:]
print('just+the+values+of+row+number+7',ser_singleRow7,+sep='\n',+end='\n\n\n')

print('='*30%2B'+JUST+FOR+COMPLETENESS+'%2B'='*30,+end='\n\n\n')

lst_fromSer=ser_singleRow7.tolist()
print('you+can+convert+a+series+to+a+list',lst_fromSer,+type(lst_fromSer),+sep='\n',+end='\n\n\n')

lst_fromDf=df.values.tolist()
print('you+can+convert+a+dataframe+to+a+nested+list',lst_fromDf,+type(lst_fromDf),+sep='\n',+end='\n\n')

arr_fromDf=df.values
print('the+content+of+a+dataframe+can+be+accessed+as+a+numpy.ndarray',arr_fromDf,+type(arr_fromDf),+sep='\n',+end='\n\n')|4649474|正如cs95指出的那样，在see+here上，其他方法应该比pandas版本0.24中的pandas+.values属性更受欢迎。我在这里使用它，因为大多数人(到2019年)仍然会使用旧版本，它不支持新的建议。您可以使用print(pd.__version__)检查您的版本|offset|length|style|CODE|4649475|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/4909087/cs95|1|https://stackoverflow.com/a/54324513/7128154^0|0|0|0|0|0|0|0|0|1F|7|32|L|2|4|0|D|8|1|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1A|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|1B|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|1D|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|1F|8|@]|9|@]|A|$I|J]]|$1|S|3|T|5|6|7|1G|8|@$U|1H|V|1I|W|X]|$U|1J|V|1K|W|X]]|9|@$U|1L|V|1M|1|1N]|$U|1O|V|1P|1|1Q]]|A|$]]|$1|Y|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]|16|$5|11|12|13|A|$14|17]]]]

As this question attained a lot of attention and there are several ways to fulfill your task, let me present several options. 

Those are all one-liners by the way ;)

Starting with:

<pre><code>df
 cluster load_date budget actual fixed_price
0 A 1/1/2014 1000 4000 Y
1 A 2/1/2014 12000 10000 Y
2 A 3/1/2014 36000 2000 Y
3 B 4/1/2014 15000 10000 N
4 B 4/1/2014 12000 11500 N
5 B 4/1/2014 90000 11000 N
6 C 7/1/2014 22000 18000 N
7 C 8/1/2014 30000 28960 N
8 C 9/1/2014 53000 51200 N
</code></pre>

Overview of potential operations:

<pre><code>ser_aggCol (collapse each column to a list)
cluster [A, A, A, B, B, B, C, C, C]
load_date [1/1/2014, 2/1/2014, 3/1/2...
budget [1000, 12000, 36000, 15000...
actual [4000, 10000, 2000, 10000,...
fixed_price [Y, Y, Y, N, N, N, N, N, N]
dtype: object


ser_aggRows (collapse each row to a list)
0 [A, 1/1/2014, 1000, 4000, Y]
1 [A, 2/1/2014, 12000, 10000...
2 [A, 3/1/2014, 36000, 2000, Y]
3 [B, 4/1/2014, 15000, 10000...
4 [B, 4/1/2014, 12000, 11500...
5 [B, 4/1/2014, 90000, 11000...
6 [C, 7/1/2014, 22000, 18000...
7 [C, 8/1/2014, 30000, 28960...
8 [C, 9/1/2014, 53000, 51200...
dtype: object


df_gr (here you get lists for each cluster)
 load_date budget actual fixed_price
cluster 
A [1/1/2014, 2/1/2014, 3/1/2... [1000, 12000, 36000] [4000, 10000, 2000] [Y, Y, Y]
B [4/1/2014, 4/1/2014, 4/1/2... [15000, 12000, 90000] [10000, 11500, 11000] [N, N, N]
C [7/1/2014, 8/1/2014, 9/1/2... [22000, 30000, 53000] [18000, 28960, 51200] [N, N, N]


a list of separate dataframes for each cluster

df for cluster A
 cluster load_date budget actual fixed_price
0 A 1/1/2014 1000 4000 Y
1 A 2/1/2014 12000 10000 Y
2 A 3/1/2014 36000 2000 Y

df for cluster B
 cluster load_date budget actual fixed_price
3 B 4/1/2014 15000 10000 N
4 B 4/1/2014 12000 11500 N
5 B 4/1/2014 90000 11000 N

df for cluster C
 cluster load_date budget actual fixed_price
6 C 7/1/2014 22000 18000 N
7 C 8/1/2014 30000 28960 N
8 C 9/1/2014 53000 51200 N

just the values of column load_date
0 1/1/2014
1 2/1/2014
2 3/1/2014
3 4/1/2014
4 4/1/2014
5 4/1/2014
6 7/1/2014
7 8/1/2014
8 9/1/2014
Name: load_date, dtype: object


just the values of column number 2
0 1000
1 12000
2 36000
3 15000
4 12000
5 90000
6 22000
7 30000
8 53000
Name: budget, dtype: object


just the values of row number 7
cluster C
load_date 8/1/2014
budget 30000
actual 28960
fixed_price N
Name: 7, dtype: object


============================== JUST FOR COMPLETENESS ==============================


you can convert a series to a list
['C', '8/1/2014', '30000', '28960', 'N']
&lt;class 'list'&gt;


you can convert a dataframe to a nested list
[['A', '1/1/2014', '1000', '4000', 'Y'], ['A', '2/1/2014', '12000', '10000', 'Y'], ['A', '3/1/2014', '36000', '2000', 'Y'], ['B', '4/1/2014', '15000', '10000', 'N'], ['B', '4/1/2014', '12000', '11500', 'N'], ['B', '4/1/2014', '90000', '11000', 'N'], ['C', '7/1/2014', '22000', '18000', 'N'], ['C', '8/1/2014', '30000', '28960', 'N'], ['C', '9/1/2014', '53000', '51200', 'N']]
&lt;class 'list'&gt;

the content of a dataframe can be accessed as a numpy.ndarray
[['A' '1/1/2014' '1000' '4000' 'Y']
 ['A' '2/1/2014' '12000' '10000' 'Y']
 ['A' '3/1/2014' '36000' '2000' 'Y']
 ['B' '4/1/2014' '15000' '10000' 'N']
 ['B' '4/1/2014' '12000' '11500' 'N']
 ['B' '4/1/2014' '90000' '11000' 'N']
 ['C' '7/1/2014' '22000' '18000' 'N']
 ['C' '8/1/2014' '30000' '28960' 'N']
 ['C' '9/1/2014' '53000' '51200' 'N']]
&lt;class 'numpy.ndarray'&gt;
</code></pre>

code:

<pre><code># prefix ser refers to pd.Series object
# prefix df refers to pd.DataFrame object
# prefix lst refers to list object

import pandas as pd
import numpy as np

df=pd.DataFrame([
 ['A', '1/1/2014', '1000', '4000', 'Y'],
 ['A', '2/1/2014', '12000', '10000', 'Y'],
 ['A', '3/1/2014', '36000', '2000', 'Y'],
 ['B', '4/1/2014', '15000', '10000', 'N'],
 ['B', '4/1/2014', '12000', '11500', 'N'],
 ['B', '4/1/2014', '90000', '11000', 'N'],
 ['C', '7/1/2014', '22000', '18000', 'N'],
 ['C', '8/1/2014', '30000', '28960', 'N'],
 ['C', '9/1/2014', '53000', '51200', 'N']
 ], columns=['cluster', 'load_date', 'budget', 'actual', 'fixed_price'])
print('df',df, sep='\n', end='\n\n')

ser_aggCol=df.aggregate(lambda x: [x.tolist()], axis=0).map(lambda x:x[0])
print('ser_aggCol (collapse each column to a list)',ser_aggCol, sep='\n', end='\n\n\n')

ser_aggRows=pd.Series(df.values.tolist()) 
print('ser_aggRows (collapse each row to a list)',ser_aggRows, sep='\n', end='\n\n\n')

df_gr=df.groupby('cluster').agg(lambda x: list(x))
print('df_gr (here you get lists for each cluster)',df_gr, sep='\n', end='\n\n\n')

lst_dfFiltGr=[ df.loc[df['cluster']==val,:] for val in df['cluster'].unique() ]
print('a list of separate dataframes for each cluster', sep='\n', end='\n\n')
for dfTmp in lst_dfFiltGr:
 print('df for cluster '+str(dfTmp.loc[dfTmp.index[0],'cluster']),dfTmp, sep='\n', end='\n\n')

ser_singleColLD=df.loc[:,'load_date']
print('just the values of column load_date',ser_singleColLD, sep='\n', end='\n\n\n')

ser_singleCol2=df.iloc[:,2]
print('just the values of column number 2',ser_singleCol2, sep='\n', end='\n\n\n')

ser_singleRow7=df.iloc[7,:]
print('just the values of row number 7',ser_singleRow7, sep='\n', end='\n\n\n')

print('='*30+' JUST FOR COMPLETENESS '+'='*30, end='\n\n\n')

lst_fromSer=ser_singleRow7.tolist()
print('you can convert a series to a list',lst_fromSer, type(lst_fromSer), sep='\n', end='\n\n\n')

lst_fromDf=df.values.tolist()
print('you can convert a dataframe to a nested list',lst_fromDf, type(lst_fromDf), sep='\n', end='\n\n')

arr_fromDf=df.values
print('the content of a dataframe can be accessed as a numpy.ndarray',arr_fromDf, type(arr_fromDf), sep='\n', end='\n\n')
</code></pre>

as pointed out by <a href="https://stackoverflow.com/users/4909087/cs95">cs95</a> other methods should be preferred over pandas <code>.values</code> attribute from pandas version 0.24 on <a href="https://stackoverflow.com/a/54324513/7128154">see here</a>. I use it here, because most people will (by 2019) still have an older version, which does not support the new recommendations. You can check your version with <code>print(pd.__version__)</code>

blocks|key|142916|text|+amount+=+list()
++++for+col+in+df.columns:
++++++++val+=+list(df[col])
++++++++for+v+in+val:
++++++++++++amount.append(v)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|142917|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> amount = list()
 for col in df.columns:
 val = list(df[col])
 for v in val:
 amount.append(v)
</code></pre>

blocks|key|142941|text|如果你的列只有一个值，像pd.series.tolist()这样的东西将产生一个错误。要保证它在所有情况下都能正常工作，请使用以下代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|142942|(
++++df
++++++++.filter(['column_name'])
++++++++.values
++++++++.reshape(1,+-1)
++++++++.ravel()
++++++++.tolist()
)|code-block|syntax|javascript|142943|entityMap^0|C|I|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If your column will only have one value something like <code>pd.series.tolist()</code> will produce an error. To guarantee that it will work for all cases, use the code below:

<pre><code>(
 df
 .filter(['column_name'])
 .values
 .reshape(1, -1)
 .ravel()
 .tolist()
)
</code></pre>

I have a dataframe <code>df</code> imported from an Excel document like this:
<pre><code>cluster load_date budget actual fixed_price
A 1/1/2014 1000 4000 Y
A 2/1/2014 12000 10000 Y
A 3/1/2014 36000 2000 Y
B 4/1/2014 15000 10000 N
B 4/1/2014 12000 11500 N
B 4/1/2014 90000 11000 N
C 7/1/2014 22000 18000 N
C 8/1/2014 30000 28960 N
C 9/1/2014 53000 51200 N
</code></pre>
I want to be able to return the contents of column 1 <code>df['cluster']</code> as a list, so I can run a for-loop over it, and create an Excel worksheet for every cluster.
Is it also possible to return the contents of a whole column or row to a list? e.g.
<pre><code>list = [], list[column1] or list[df.ix(row1)]
</code></pre>

Get list from pandas dataframe column or row?

Python

我有一个从Excel文档导入的dataframe df，如下所示：cluster load_date   budget  actual  fixed_priceA   1/1/2014    1000    4000    YA   2/1/2014    12000   10000   YA   3/1/2014  ...

问从pandas dataframe列或行中获取列表？
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