blocks|key|4072227|text|只需查找非空项，然后对各行求和：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4072228|df['Chains']+=+df.notnull().sum(axis=1)+-+1|code-block|syntax|javascript|4072229|我不得不减去一个来计算你的IndividualID专栏。这是我得到的结果：|offset|length|style|CODE|4072230|+++IndividualID++Trip1++Trip2++Trip3++Trip4++Trip5++Trip6++Trip7++Trip8++Trip9++Chains
0+++++200100001+++++23++++1.0++++2.0++++4.0++++4.0++++1.0++++5.0++++5.0++++5.0+++++++9
1+++++200100002+++++21++++1.0+++12.0++++3.0++++1.0+++55.0++++7.0++++7.0++++NaN+++++++8
2+++++200100003+++++12++++3.0++++3.0++++6.0++++3.0++++NaN++++NaN++++NaN++++NaN+++++++5
3+++++200100004++++++4++++NaN++++NaN++++NaN++++NaN++++NaN++++NaN++++NaN++++NaN+++++++1
4+++++200100005++++++6++++5.0++++3.0++++9.0++++3.0++++5.0++++6.0++++NaN++++NaN+++++++7
5+++++200100005+++++23++++4.0++++4.0++++2.0++++4.0++++3.0++++6.0++++5.0++++NaN+++++++8|4072231|entityMap^0|0|0|D|C|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Simply look for the non-null items and then sum the rows:

<pre><code>df['Chains'] = df.notnull().sum(axis=1) - 1
</code></pre>

I had to subtract one to account for your <code>IndividualID</code> column. This is the result I got:

<pre><code> IndividualID Trip1 Trip2 Trip3 Trip4 Trip5 Trip6 Trip7 Trip8 Trip9 Chains
0 200100001 23 1.0 2.0 4.0 4.0 1.0 5.0 5.0 5.0 9
1 200100002 21 1.0 12.0 3.0 1.0 55.0 7.0 7.0 NaN 8
2 200100003 12 3.0 3.0 6.0 3.0 NaN NaN NaN NaN 5
3 200100004 4 NaN NaN NaN NaN NaN NaN NaN NaN 1
4 200100005 6 5.0 3.0 9.0 3.0 5.0 6.0 NaN NaN 7
5 200100005 23 4.0 4.0 2.0 4.0 3.0 6.0 5.0 NaN 8
</code></pre>

blocks|key|4073724|text|将所有空值替换为NaN，然后使用sum(1)逐行计算notnull值|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4073725|df['Chains']+=+df.iloc[:,1:].replace('',np.nan).notnull().sum(1)

>>>+df
+++IndividualID++Trip1++Trip2++Trip3++Trip4++Trip5++Trip6++Trip7++Trip8++\
0+++++200100001+++++23++++1.0++++2.0++++4.0++++4.0++++1.0++++5.0++++5.0+++
1+++++200100002+++++21++++1.0+++12.0++++3.0++++1.0+++55.0++++7.0++++7.0+++
2+++++200100003+++++12++++3.0++++3.0++++6.0++++3.0++++NaN++++NaN++++NaN+++
3+++++200100004++++++4++++NaN++++NaN++++NaN++++NaN++++NaN++++NaN++++NaN+++
4+++++200100005++++++6++++5.0++++3.0++++9.0++++3.0++++5.0++++6.0++++NaN+++
5+++++200100005+++++23++++4.0++++4.0++++2.0++++4.0++++3.0++++6.0++++5.0+++

+++Trip9++Chains++
0++++5.0+++++++9++
1++++NaN+++++++8++
2++++NaN+++++++5++
3++++NaN+++++++1++
4++++NaN+++++++7++
5++++NaN+++++++8++|code-block|syntax|javascript|4073726|entityMap^0|8|3|G|6|Q|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Replace all the blank values to <code>NaN</code>, then count the <code>notnull</code> values by row using <code>sum(1)</code>:

<pre><code>df['Chains'] = df.iloc[:,1:].replace('',np.nan).notnull().sum(1)

&gt;&gt;&gt; df
 IndividualID Trip1 Trip2 Trip3 Trip4 Trip5 Trip6 Trip7 Trip8 \
0 200100001 23 1.0 2.0 4.0 4.0 1.0 5.0 5.0 
1 200100002 21 1.0 12.0 3.0 1.0 55.0 7.0 7.0 
2 200100003 12 3.0 3.0 6.0 3.0 NaN NaN NaN 
3 200100004 4 NaN NaN NaN NaN NaN NaN NaN 
4 200100005 6 5.0 3.0 9.0 3.0 5.0 6.0 NaN 
5 200100005 23 4.0 4.0 2.0 4.0 3.0 6.0 5.0 

 Trip9 Chains 
0 5.0 9 
1 NaN 8 
2 NaN 5 
3 NaN 1 
4 NaN 7 
5 NaN 8 
</code></pre>

blocks|key|4073753|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4073754|df.ne('').sum(1)-1
Out[287]:+
0++++9
1++++8
2++++5
3++++1
4++++7
5++++8
dtype:+int64|code-block|syntax|javascript|4073755|如果是使用info的NaN|offset|length|style|CODE|4073756|df.iloc[:,1:].T.info()
<class+'pandas.core.frame.DataFrame'>
Index:+9+entries,+Trip1+to+Trip9
Data+columns+(total+6+columns):
0++++9+non-null+float64
1++++8+non-null+float64
2++++5+non-null+float64
3++++1+non-null+float64
4++++7+non-null+float64
5++++8+non-null+float64
dtypes:+float64(6)
memory+usage:+504.0%2B+bytes|4073757|entityMap^0|0|0|5|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Using 

<pre><code>df.ne('').sum(1)-1
Out[287]: 
0 9
1 8
2 5
3 1
4 7
5 8
dtype: int64
</code></pre>

If it is NaN using <code>info</code>

<pre><code>df.iloc[:,1:].T.info()
&lt;class 'pandas.core.frame.DataFrame'&gt;
Index: 9 entries, Trip1 to Trip9
Data columns (total 6 columns):
0 9 non-null float64
1 8 non-null float64
2 5 non-null float64
3 1 non-null float64
4 7 non-null float64
5 8 non-null float64
dtypes: float64(6)
memory usage: 504.0+ bytes
</code></pre>

blocks|key|4072315|text|也许：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4072316|>>>+df.replace('',pd.np.nan).count(axis=1)-1
0++++9
1++++8
2++++5
3++++1
4++++7
5++++8
dtype:+int64|code-block|syntax|javascript|4072317|或者，如果有nan这样做：|offset|length|style|CODE|4072318|>>>+df.count(axis=1)-1
0++++9
1++++8
2++++5
3++++1
4++++7
5++++8
dtype:+int64|4072319|然后就这么做：|4072320|df['Chains']+=+...|4072321|用于将其分配给列|4072322|entityMap^0|0|0|6|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Maybe:

<pre><code>&gt;&gt;&gt; df.replace('',pd.np.nan).count(axis=1)-1
0 9
1 8
2 5
3 1
4 7
5 8
dtype: int64
</code></pre>

Or if there <code>nan</code> do:

<pre><code>&gt;&gt;&gt; df.count(axis=1)-1
0 9
1 8
2 5
3 1
4 7
5 8
dtype: int64
</code></pre>

And just do:

<pre><code>df['Chains'] = ...
</code></pre>

For assigning it to a column

blocks|key|4072348|text|只要我们提供备选方案，如果值为NaN|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4072349|df['cat']+=+(~np.isnan(df.set_index('IndividualID').values)).sum(1)

IndividualID
200100001++++9
200100002++++8
200100003++++5
200100004++++1
200100005++++7
200100005++++8|code-block|syntax|javascript|4072350|entityMap^0|F|3|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|$]]

So long as we're giving alternatives, if values are <code>NaN</code>

<pre><code>df['cat'] = (~np.isnan(df.set_index('IndividualID').values)).sum(1)

IndividualID
200100001 9
200100002 8
200100003 5
200100004 1
200100005 7
200100005 8
</code></pre>

I have a dataframe like this:

<pre><code>IndividualID Trip1 Trip2 Trip3 Trip4 Trip5 Trip6 Trip7 Trip8 Trip9
200100001 23 1 2 4 4 1 5 5 5
200100002 21 1 12 3 1 55 7 7
200100003 12 3 3 6 3 
200100004 4 
200100005 6 5 3 9 3 5 6 
200100005 23 4 4 2 4 3 6 5 
</code></pre>

I'm trying to know the number of trips that each individual makes, so I would like to create a new column so the new table would probably look like this: 

<pre><code>IndividualID Trip1 Trip2 Trip3 Trip4 Trip5 Trip6 Trip7 Trip8 Trip9 Chains
200100001 23 1 2 4 4 1 5 5 5 9
200100002 21 1 12 3 1 55 7 7 8
200100003 12 3 3 6 3 5
200100004 4 1
200100005 6 5 3 9 3 5 6 7
200100005 23 4 4 2 4 3 6 5 8
</code></pre>

Are there any possible solutions? I would really appreciate if someone can help with it! Thanks in advance!

How to count the number of columns with a value on each row in python?

我有一个这样的数据帧：IndividualID Trip1 Trip2 Trip3 Trip4 Trip5 Trip6 Trip7 Trip8 Trip9200100001    23    1     2     4     4      1    5     5     5200100002    21    1 ...

问如何在python中统计每行有值的列数？
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